🧠:[<think>Okay, so I have this quadratic equation: kx² - 2(3k - 1)x + 9k - 1 = 0, and I need to find all real numbers k such that this equation has two integer roots. Hmm, let's think step by step.First, since it's a quadratic equation in x, the coefficient of x², which is k, can't be zero. Otherwise, it would become a linear equation. So k ≠ 0. Got that.Now, for a quadratic equation to have two real roots, the discriminant must be non-negative. But in this case, the problem says the equation has two integer roots. So not just real, but integers. Therefore, first, the discriminant must be a perfect square because the roots are integers. Also, the roots have to actually be integers. Let's recall that for a quadratic equation ax² + bx + c = 0, the roots are given by (-b ± √D)/(2a), where D is the discriminant.So let's compute the discriminant D for this equation. The equation is kx² - 2(3k - 1)x + (9k - 1) = 0. Therefore, a = k, b = -2(3k - 1), c = 9k - 1.Discriminant D = b² - 4ac = [ -2(3k - 1) ]² - 4 * k * (9k - 1).Let me compute that step by step. First, square of -2(3k - 1):[-2(3k - 1)]² = 4(3k - 1)² = 4*(9k² - 6k + 1) = 36k² - 24k + 4.Then, subtract 4ac: 4 * k * (9k - 1) = 4k*(9k - 1) = 36k² - 4k.So D = (36k² -24k +4) - (36k² -4k) = 36k² -24k +4 -36k² +4k = (-20k) +4.Therefore, discriminant D = -20k + 4. Wait, that seems a bit strange. Let me check again:Wait, 36k² -24k +4 minus 36k² +4k. So 36k² cancels with -36k². Then -24k +4k is -20k. Then +4 remains. Yes, so D = -20k +4. But for the quadratic to have real roots, D must be non-negative. So -20k +4 ≥0 ⇒ -20k ≥ -4 ⇒ 20k ≤4 ⇒ k ≤ 4/20 ⇒ k ≤ 1/5.But we also need that D is a perfect square because the roots are integers. So D must be a perfect square, and the roots must be integers. So let's note that D must be a perfect square. Let me denote D = m² where m is a non-negative integer because D is a square.So m² = -20k +4 ⇒ 20k = 4 -m² ⇒ k = (4 -m²)/20.But k must be a real number (though here since m² is an integer, k would be rational if m is integer). Wait, but m is a non-negative integer such that D = m² is non-negative, so m² = -20k +4 ≥0, so as before.So k = (4 -m²)/20. So m² must be less than or equal to 4, since k must be real (and m is a non-negative integer). Therefore, m² ≤4 ⇒ m ∈ {0,1,2}.So m can be 0,1,2. Let's check each case.Case 1: m=0. Then D=0, so the equation has a repeated root. So the root is [-b/(2a)]. Since the problem says "two integer roots", but if discriminant is zero, there's only one root (a double root). So does the problem allow for a double root? The problem says "two integer roots". Hmm, depending on interpretation, sometimes "two" can mean two distinct, but sometimes it can include multiplicity. But in the problem statement, it's better to check both possibilities. Let's see.But first, let's proceed. If m=0, D=0. Then k=(4 -0)/20=4/20=1/5. Then check if the equation has a double root which is integer. So let's compute the root.The equation is kx² -2(3k -1)x +9k -1=0. If k=1/5, then substitute:(1/5)x² -2(3*(1/5) -1)x +9*(1/5) -1=0Compute coefficients:First term: (1/5)x²Second term: -2*(3/5 -1) = -2*(-2/5)= 4/5, so second term is (4/5)xThird term: 9/5 -1 = 4/5So equation becomes (1/5)x² + (4/5)x +4/5=0. Multiply both sides by 5:x² +4x +4=0 ⇒ (x+2)²=0. So root is x=-2, double root. So the equation has two integer roots (both -2). If the problem counts multiplicity, then k=1/5 is acceptable. But if the problem requires two distinct integer roots, then m=0 (k=1/5) would be invalid. Since the problem says "two integer roots", but doesn't specify distinct, so maybe we have to include k=1/5. Let's keep it for now and check later.Case 2: m=1. Then D=1. So k=(4 -1)/20=3/20. Then check if the equation has two integer roots. Let's compute the roots.The quadratic equation is kx² -2(3k -1)x +9k -1=0. Substitute k=3/20:First term: (3/20)x²Second term: -2*(3*(3/20) -1) = -2*(9/20 -1) = -2*(-11/20) = 22/20 = 11/10. So second term is (11/10)x.Third term: 9*(3/20) -1 = 27/20 -20/20 =7/20.So equation is (3/20)x² + (11/10)x +7/20=0. Multiply both sides by 20 to eliminate denominators:3x² +22x +7=0.Now, compute roots: x = [-22 ± √(22² -4*3*7)]/(2*3) = [-22 ± √(484 -84)]/6 = [-22 ± √400]/6 = [-22 ±20]/6.So two roots:(-22 +20)/6 = (-2)/6 = -1/3.(-22 -20)/6 = (-42)/6 = -7.So one root is -7 (integer), and another is -1/3 (not integer). Therefore, m=1 gives one integer root and one non-integer root. So this case is invalid. Therefore, k=3/20 is rejected.Case 3: m=2. Then D=4. So k=(4 -4)/20=0/20=0. But earlier we said k cannot be zero, since the equation would not be quadratic anymore. So this is invalid. So m=2 gives k=0, which is excluded.Wait, but m=2 leads to k=0, which is not allowed. Therefore, m=2 is invalid.So from the discriminant analysis, the possible m values are 0 and 1. But m=1 gives k=3/20 which leads to one integer root and one non-integer root, so invalid. m=0 gives k=1/5, which gives a double root at x=-2, which is integer. So k=1/5 is a candidate.But wait, the problem says "two integer roots". If it requires two distinct integer roots, then k=1/5 is invalid because there's only one root. If it allows a repeated root, then k=1/5 is valid. So we need to check the problem statement.The original problem says "has two integer roots". In standard terminology, sometimes "two" can mean two roots counting multiplicity, but sometimes it might mean two distinct roots. Since the problem didn't specify "distinct", but in quadratic equations, unless specified, having two roots can include the case of a double root. However, sometimes problems might mean two distinct roots. So maybe we need to check both possibilities. But in the discriminant approach, we considered D as a perfect square, and found k=1/5. Let's check if there are other possibilities.Wait, perhaps there's another approach. Let's consider that if the quadratic equation has two integer roots, then we can express the quadratic in factored form as k(x - p)(x - q) = 0, where p and q are integers. Then expanding this:k x² -k(p + q)x +kpq =0.Comparing coefficients with the given equation:Given equation: kx² -2(3k -1)x +9k -1=0.Therefore, equate coefficients:1. Coefficient of x²: k = k (same on both sides)2. Coefficient of x: -k(p + q) = -2(3k -1)3. Constant term: kpq =9k -1So from the coefficient of x:-k(p + q) = -2(3k -1) ⇒ k(p + q) = 2(3k -1) ⇒ k(p + q) =6k -2 ⇒ k(p + q -6) = -2 ⇒ k = -2/(p + q -6)From the constant term:kpq =9k -1 ⇒ k(pq -9) = -1 ⇒ k = -1/(pq -9)So we have two expressions for k:k = -2/(p + q -6) and k = -1/(pq -9)Therefore, set them equal:-2/(p + q -6) = -1/(pq -9) ⇒ 2/(p + q -6) =1/(pq -9) ⇒ Cross-multiplying: 2(pq -9) = p + q -6 ⇒ 2pq -18 = p + q -6 ⇒ 2pq -p -q =12 ⇒ 2pq -p -q =12.So we have the equation 2pq - p - q =12. Let's try to factor this or find integer solutions.Let me rearrange:2pq -p - q =12. Let's add 1 to both sides:2pq -p - q +1 =13. Then factor left side:(2pq -p - q +1) = (2p -1)(q -1) +0? Wait, maybe try to factor.Alternatively, add 1 to both sides: 2pq - p - q +1 =13 ⇒ (2p -1)(2q -1)/2 =13? Not sure. Alternatively, use Simon's Favorite Factoring Trick.Let me write 2pq -p - q =12.Multiply both sides by 1: 2pq -p - q =12. Let me factor:p(2q -1) - q =12. Hmm, not obvious.Alternatively, write as:2pq - p - q =12 ⇒ 4pq -2p -2q =24 ⇒ (2p -1)(2q -1) -1 =24 ⇒ (2p -1)(2q -1)=25.Ah, that works. Let's verify:Start from 2pq - p - q =12.Multiply both sides by 2: 4pq -2p -2q =24.Add 1 to both sides: 4pq -2p -2q +1 =25.Factor the left-hand side: (2p -1)(2q -1) =25.Yes! Because (2p -1)(2q -1) =4pq -2p -2q +1. Perfect. So:(2p -1)(2q -1)=25.Since p and q are integers, 2p -1 and 2q -1 must be integers that multiply to 25. The factors of 25 are:1 and 25,5 and 5,-1 and -25,-5 and -5.So possible pairs (a,b) where a*b=25:(1,25), (25,1), (5,5), (-1,-25), (-25,-1), (-5,-5).Therefore, for each such pair, set:2p -1 = a and 2q -1 = b. Then solve for p and q.Let's go through each possible pair.Case 1: (a,b) = (1,25)Then 2p -1 =1 ⇒ 2p=2 ⇒ p=1.2q -1=25 ⇒2q=26 ⇒ q=13.So p=1, q=13. Check if these are integers. Yes.Compute k using k = -2/(p + q -6). p+q=1+13=14. So 14-6=8. So k= -2/8= -1/4.But we also have k = -1/(pq -9). pq=1*13=13. pq -9=4. So k= -1/4. Same result. So k= -1/4.Check if the quadratic equation with k= -1/4 has two integer roots.Equation: (-1/4)x² -2(3*(-1/4) -1)x +9*(-1/4) -1=0.Simplify coefficients:First term: (-1/4)x².Second term: -2*( -3/4 -1 )x = -2*(-7/4)x = (14/4)x = (7/2)x.Third term: -9/4 -1 = -13/4.So equation: (-1/4)x² + (7/2)x -13/4 =0. Multiply by -4 to eliminate denominators:x² -14x +13=0.Factor: (x-1)(x-13)=0. So roots at x=1 and x=13. Both integers. Perfect. So k= -1/4 is valid.Case 2: (a,b)=(25,1). This is symmetric to case 1. So 2p -1=25 ⇒ p=13, 2q -1=1 ⇒ q=1. So same as above, p=13, q=1. This gives same k=-1/4.Case3: (a,b)=(5,5). Then 2p -1=5 ⇒2p=6 ⇒p=3. Similarly, 2q -1=5 ⇒q=3. So p=q=3.Compute k= -2/(3 +3 -6)= -2/0. Division by zero, invalid. So this case is invalid.Case4: (a,b)=(-1,-25). Then 2p -1=-1 ⇒2p=0 ⇒p=0. 2q -1=-25 ⇒2q=-24 ⇒q=-12. So p=0, q=-12.Compute k= -2/(0 + (-12) -6)= -2/(-18)=1/9.Also check with k= -1/(pq -9). pq=0*(-12)=0. pq -9= -9. So k= -1/(-9)=1/9. Same result.Check the quadratic equation with k=1/9.Equation: (1/9)x² -2(3*(1/9) -1)x +9*(1/9) -1=0.Simplify:First term: (1/9)x².Second term: -2*(1/3 -1)x = -2*(-2/3)x = (4/3)x.Third term:1 -1=0.So equation: (1/9)x² + (4/3)x =0. Multiply by 9:x² +12x=0 ⇒x(x +12)=0. Roots at x=0 and x=-12. Both integers. So k=1/9 is valid.Case5: (a,b)=(-25,-1). Symmetric to case4. So p= -12, q=0. Same result, k=1/9.Case6: (a,b)=(-5,-5). Then 2p -1=-5 ⇒2p=-4 ⇒p=-2. Similarly, 2q -1=-5 ⇒q=-2.Compute k= -2/((-2)+(-2)-6)= -2/(-10)=1/5.Also check with k= -1/(pq -9). pq=(-2)(-2)=4. pq -9= -5. So k= -1/(-5)=1/5. Same result.Check the quadratic equation with k=1/5. Wait, we already did that earlier. It gave a double root at x=-2. So in this case, p=q=-2. So the quadratic equation would be k(x +2)^2=0, which expands to kx² +4kx +4k=0. Comparing with given equation:Given equation is kx² -2(3k -1)x +9k -1=0.So equate coefficients:k x² +4k x +4k = kx² -2(3k -1)x +9k -1Therefore, coefficients:4k = -2(3k -1) ⇒4k = -6k +2 ⇒10k=2 ⇒k=1/5. Then check constant term:4k=9k -1 ⇒4*(1/5)=9*(1/5)-1 ⇒4/5=9/5 -5/5 ⇒4/5=4/5. So yes, it works. The equation has a double root at x=-2, which is integer. So if the problem allows a double root (i.e., two integer roots with multiplicity), then k=1/5 is valid. If it requires two distinct roots, then k=1/5 is invalid.But according to the cases above from the factoring approach, we have k= -1/4, 1/9, and 1/5. But when m=0 (discriminant approach), we found k=1/5. But through factoring, we found k= -1/4, 1/9, and 1/5. So why the discrepancy? Because in discriminant approach, m was 0,1,2, but in the factoring approach, we get additional solutions.Wait, this suggests that the discriminant approach missed some possible values of k. Because when we considered m as integers 0,1,2, leading to k=1/5,3/20,0. But through the factoring approach, we found other k's. So the discriminant approach was incomplete. That’s because the discriminant must be a perfect square, but we considered m as non-negative integers. However, in reality, D must be a perfect square, but m can be any integer (positive or negative), but since D is non-negative, m is real. Wait, actually in the discriminant approach, when I set D=m², m is a real number such that m²=D. But for the roots to be integers, the discriminant must be a perfect square, yes, but m could be a positive or negative integer. However, in the previous approach, I considered m as non-negative integers from 0 to 2, but actually m can be any integer (positive or negative) as long as m²=D. However, in the equation D=-20k+4, m²=-20k +4, so m must satisfy that -20k +4 is a perfect square, but since k can be positive or negative, m² can be any non-negative integer such that m²=4 -20k. Therefore, m² must be congruent to 4 mod 20, but that's perhaps complicating.Wait, but in the factoring approach, we found k= -1/4, 1/9, and 1/5. Let's check their discriminants:For k=-1/4:D=-20*(-1/4)+4=5 +4=9=3², which is a perfect square.For k=1/9:D=-20*(1/9)+4= -20/9 +36/9=16/9=(4/3)^2. Wait, 4/3 is not integer. But the roots are integers. Wait, but if D=(4/3)^2, then sqrt(D)=4/3, so roots would be:x=[2(3k -1) ±4/3]/(2k). Wait, but in this case, k=1/9. Let me compute the roots again.Wait, when k=1/9:The equation becomes (1/9)x² -2(3*(1/9)-1)x +9*(1/9)-1=0 ⇒ (1/9)x² -2(-2/3)x +0=0 ⇒ (1/9)x² + (4/3)x=0 ⇒ Multiply by 9: x² +12x=0 ⇒x(x+12)=0. So roots x=0 and x=-12, which are integers. But discriminant D= (4/3)^2=16/9. So even though D is a square of a rational number, the roots are integers. Hmm, so in this case, the discriminant is a square of a rational number, not necessarily an integer, but when divided by 2a=2*(1/9)=2/9, the numerator and denominator combine to give integers. So the discriminant doesn't need to be a perfect square of an integer, just a perfect square in rationals. Wait, but in quadratic equations, the discriminant being a perfect square (in integers) is required for roots to be rational, but here the problem says the roots must be integers. So for roots to be integers, discriminant must be a perfect square, and the numerator in the quadratic formula must be even. Wait, maybe I made a mistake in the discriminant approach.Wait, the standard rational root theorem says that any rational root p/q (in lowest terms) of a polynomial with integer coefficients satisfies that p divides the constant term and q divides the leading coefficient. But in our problem, the equation is kx² -2(3k -1)x +9k -1=0. If k is rational, say k=a/b with a and b integers, then multiplying through by b would give a quadratic with integer coefficients. Then the roots would be integers only if they satisfy the integer root theorem for that polynomial.But maybe instead of complicating with discriminant, the approach via factoring into integer roots is better. Which we did and found k= -1/4, 1/9, and 1/5. Wait, but in discriminant approach, we thought only k=1/5 was possible. But clearly, when k=-1/4 and k=1/9, the equation has integer roots. Therefore, the discriminant approach was flawed because when k is not an integer, the discriminant can be a square of a rational number, leading to roots that are integers when combined with the denominator in the quadratic formula. So let me correct the discriminant approach.In discriminant approach:We have D = -20k +4. For roots to be integers, the roots [2(3k -1) ±√D]/(2k) must be integers.Let me denote sqrt(D) as m. So m = √(-20k +4). Then:The roots are [2(3k -1) ±m]/(2k). Let me denote the roots as p and q (integers). So:p = [2(3k -1) +m]/(2k)q = [2(3k -1) -m]/(2k)Adding these two equations: p + q = [2(3k -1) +m +2(3k -1) -m]/(2k) = [4(3k -1)]/(2k) = 2(3k -1)/k = 6 - 2/k.Similarly, multiplying p and q:pq = [2(3k -1) +m][2(3k -1) -m]/(4k²) = [4(3k -1)² -m²]/(4k²).But since m² = D = -20k +4, substitute:pq = [4(9k² -6k +1) - (-20k +4)]/(4k²) = [36k² -24k +4 +20k -4]/(4k²) = [36k² -4k]/(4k²) = (4k(9k -1))/(4k²) = (9k -1)/k.So we have:p + q =6 - 2/k,pq =9 -1/k.Since p and q are integers, 6 -2/k must be integer, and 9 -1/k must be integer.Let me denote 2/k as an integer. Let’s set 2/k = integer. Let’s let t =1/k. Then 6 -2/k =6 -2t must be integer. Similarly, 9 -1/k=9 -t must be integer. Therefore, t must be a rational number such that 6 -2t and 9 -t are integers. Since t=1/k, k ≠0.Let’s denote t =1/k. Then:6 -2t ∈ ℤ,9 -t ∈ ℤ.Let’s subtract the two equations:(9 -t) - (6 -2t)=3 +t ∈ ℤ.Therefore, 3 +t must be integer. So since 9 -t is integer, and 3 +t is integer, then t must be integer. Because let me write:Let’s set 9 -t = m ∈ ℤ,3 +t =n ∈ ℤ.Then from 9 -t =m and 3 +t =n, adding them:12 =m +n.So m and n are integers such that m +n=12.From 9 -t =m ⇒ t=9 -m,From 3 +t =n ⇒ t=n -3.Therefore, 9 -m =n -3 ⇒ m +n=12, which is consistent.Since t=1/k must be integer, so t ∈ ℤ. Therefore, t is integer.Therefore, k=1/t, where t is integer. Also, t ≠0.Additionally, from the original equations:p + q=6 -2t,pq=9 -t.Since p and q are integers, 6 -2t and 9 -t must be integers, which they are since t is integer.So t is integer, and k=1/t. So possible integer values of t are t ∈ ℤ {0}.But also, from discriminant D = -20k +4 = m² ≥0 ⇒ -20k +4 ≥0 ⇒ k ≤1/5. Since k=1/t, then 1/t ≤1/5. So if t>0, then t≥5. If t<0, then 1/t ≤1/5 ⇒ since t negative, 1/t ≤1/5 ⇒ t≥5 (but t is negative, so contradiction). Therefore, for t positive: t≥5, but 1/t ≤1/5. For t negative: 1/t ≤1/5 ⇒ since t negative, 1/t ≤1/5 ⇨ t≥5 (but t negative). Therefore, only possible when t≥5 (positive integers) or t negative.Wait, let's re-examine:If k=1/t, then k ≤1/5.If t is positive, then 1/t ≤1/5 ⇨ t ≥5.If t is negative, then 1/t ≤1/5 ⇨ since t negative, 1/t ≤1/5 is always true because 1/t is negative, which is ≤1/5. So for t negative integers, k=1/t will satisfy k ≤1/5 automatically.Therefore, possible t are:t positive integers ≥5,t negative integers.But let's check whether these t lead to integer p and q.Given p + q=6 -2t,pq=9 -t.So for each integer t (positive ≥5 or negative), we can check if 6 -2t and 9 -t lead to integer p and q which are roots. However, this might take a long time. But earlier via factoring approach, we found that k= -1/4, 1/9, and1/5. Let's see the corresponding t:For k= -1/4, t=1/k= -4.For k=1/9, t=9.For k=1/5, t=5.So these correspond to t=-4,5,9. So t=-4 (negative), t=5,9 (positive). So these values are within t ≥5 (positive) or t negative. Therefore, perhaps we can find all t such that p and q are integers.So given t is integer, t≠0, and p + q=6 -2t, pq=9 -t. So given that p and q are integers, we can write:Let p and q be integers such that their sum is S=6 -2t and product P=9 -t.Therefore, for each t, we can check if the equations x² - Sx + P=0 have integer roots. So the quadratic equation x² - (6 -2t)x + (9 -t)=0 must have integer roots. Therefore, discriminant of this quadratic in x must be a perfect square.Compute discriminant D’= [6 -2t]^2 -4*(9 -t) =36 -24t +4t² -36 +4t=4t² -20t.Therefore, D’=4t² -20t=4t(t -5). For D’ to be a perfect square, 4t(t -5) must be a perfect square. Let’s factor:4t(t -5)= [2√t(t -5)]². Therefore, t(t -5) must be a perfect square.So t(t -5) must be a perfect square. Let me denote t(t -5)=n², where n is a non-negative integer.So equation: t(t -5)=n².We need to solve this Diophantine equation. Let’s rearrange:t² -5t -n²=0.Treat this as quadratic in t:t = [5 ±√(25 +4n²)]/2.For t to be integer, 25 +4n² must be a perfect square. Let’s set 25 +4n²=m², where m is a positive integer.So m² -4n²=25. This is a difference of squares: (m -2n)(m +2n)=25.25 can be factored into 1*25, 5*5, (-1)*(-25), (-5)*(-5). Since m and n are positive integers (since n² is non-negative), m must be greater than 2n.So possible factor pairs:(m -2n, m +2n)=(1,25), (5,5). Negative factors would lead to negative m or n, which we can disregard since m and n are positive.Case1: (1,25). Then:m -2n=1,m +2n=25.Adding equations: 2m=26 ⇒m=13. Subtracting:4n=24 ⇒n=6.Therefore, t=(5 ±m)/2. From earlier, t=(5 ±m)/2. Wait, but m=13:t=(5 ±13)/2. So t=(5 +13)/2=18/2=9, or t=(5 -13)/2=(-8)/2=-4.So t=9 or t=-4.Check if these satisfy t(t -5)=n².For t=9: 9*(9-5)=9*4=36=6². Yes.For t=-4: (-4)*(-4 -5)=(-4)*(-9)=36=6². Yes.Case2: (5,5). Then:m -2n=5,m +2n=5.Adding:2m=10 ⇒m=5. Subtracting:4n=0 ⇒n=0. But n=0 would lead to t(t -5)=0 ⇒t=0 or t=5. But t=0 is invalid because k=1/t undefined, and t=5:Check t=5:5*0=0=0², but n=0. But n must be non-negative integer, so n=0 is allowed. So t=5 or t=0. t=0 is invalid, t=5.So t=5: Check t(t -5)=5*0=0=0². Yes. So n=0. So possible t=5.Thus, solutions are t=9, t=-4, t=5.So t=9: Then k=1/9. As before.t=-4: k= -1/4.t=5: k=1/5.These are the only solutions. Hence, the values of k are k=1/5, k=1/9, and k=-1/4.Thus, the real numbers k are -1/4, 1/9, and 1/5.Wait, but when t=5, check p and q:p + q=6 -2*5=6 -10=-4,pq=9 -5=4.So integers p and q such that p + q=-4 and pq=4. The roots would be -2 and -2. So double root at -2. Which we saw earlier. So yes, that's valid.Therefore, the solutions are k= -1/4, 1/9, and 1/5. These are the three values of k for which the quadratic equation has two integer roots (counting multiplicity in the case of k=1/5).So the answer is k= -1/4, 1/9, and 1/5. Let me verify once more.For k=-1/4: roots 1 and13? Wait, no, earlier when k=-1/4, the equation became x² -14x +13=0, roots 1 and13. Yes.For k=1/9: roots 0 and -12.For k=1/5: double root at -2.All integers. So these are valid.Hence, the real numbers k are -1/4, 1/9, and 1/5.Final AnswerThe real numbers ( k ) are (boxed{-dfrac{1}{4}}), (boxed{dfrac{1}{9}}), and (boxed{dfrac{1}{5}}).</think>To find all real numbers ( k ) such that the quadratic equation ( kx^2 - 2(3k - 1)x + 9k - 1 = 0 ) has two integer roots, we start by considering the discriminant and the factored form of the quadratic equation.1. Discriminant Analysis: - The discriminant ( D ) of the quadratic equation ( kx^2 - 2(3k - 1)x + 9k - 1 = 0 ) is given by: [ D = [-2(3k - 1)]^2 - 4k(9k - 1) = 36k^2 - 24k + 4 - 36k^2 + 4k = -20k + 4 ] - For the quadratic to have real roots, ( D geq 0 ), which gives ( k leq frac{1}{5} ). For the roots to be integers, ( D ) must be a perfect square.2. Factored Form Approach: - Express the quadratic equation in factored form ( k(x - p)(x - q) = 0 ) where ( p ) and ( q ) are integers. Equate coefficients to get: [ k(p + q) = 2(3k - 1) quad text{and} quad kpq = 9k - 1 ] - Solving these, we find: [ k = frac{-2}{p + q - 6} quad text{and} quad k = frac{-1}{pq - 9} ] - Equating these expressions for ( k ), we derive the equation: [ 2pq - p - q = 12 ] - Using Simon's Favorite Factoring Trick, we factorize and solve for integer pairs ( (p, q) ): [ (2p - 1)(2q - 1) = 25 ] - Possible factor pairs of 25 lead to solutions for ( p ) and ( q ), and subsequently for ( k ).3. Solutions: - Solving the factor pairs, we find the possible values of ( k ): - For factor pairs ( (1, 25) ) and ( (25, 1) ), we get ( k = -frac{1}{4} ). - For factor pairs ( (-1, -25) ) and ( (-25, -1) ), we get ( k = frac{1}{9} ). - For factor pair ( (-5, -5) ), we get ( k = frac{1}{5} ).4. Verification: - Check each ( k ) to ensure the quadratic equation has integer roots: - ( k = -frac{1}{4} ) gives roots 1 and 13. - ( k = frac{1}{9} ) gives roots 0 and -12. - ( k = frac{1}{5} ) gives a double root -2.Thus, the real numbers ( k ) are (boxed{-dfrac{1}{4}}), (boxed{dfrac{1}{9}}), and (boxed{dfrac{1}{5}}).,<think>Okay, let's see. I need to find all real numbers k such that the quadratic equation kx² - 2(3k - 1)x + 9k - 1 = 0 has two integer roots. Hmm, quadratic equations with integer roots. Let me recall some concepts.First, for a quadratic equation ax² + bx + c = 0, the roots are given by the quadratic formula: [-b ± √(b² - 4ac)]/(2a). For the roots to be integers, several conditions must be met. The discriminant D = b² - 4ac must be a perfect square because the square root of D needs to be an integer for the roots to be rational (and since they need to be integers, even more so). Also, once we have the roots expressed as fractions, their denominators should divide the numerator to result in integers.But maybe there's another approach using Vieta's formulas. If the equation has two integer roots, say p and q, then according to Vieta's:p + q = 2(3k - 1)/kandpq = (9k - 1)/kSince p and q are integers, the expressions (2(3k - 1))/k and (9k - 1)/k must also be integers. Let me write that down:1. (6k - 2)/k must be an integer. Simplify this: 6 - 2/k must be an integer.2. (9k - 1)/k must be an integer. Simplify this: 9 - 1/k must be an integer.So both 6 - 2/k and 9 - 1/k are integers. Let's denote 6 - 2/k = m and 9 - 1/k = n, where m and n are integers. Then:From the first equation: 6 - m = 2/k ⇒ k = 2/(6 - m)From the second equation: 9 - n = 1/k ⇒ k = 1/(9 - n)Therefore, 2/(6 - m) = 1/(9 - n) ⇒ 2(9 - n) = 6 - m ⇒ 18 - 2n = 6 - m ⇒ m = 2n - 12But since m and n are integers, and k must be a real number (except possibly k ≠ 0 because it's a quadratic equation), so denominators can't be zero. Let's see:From k = 2/(6 - m), 6 - m ≠ 0 ⇒ m ≠ 6Similarly, from k = 1/(9 - n), 9 - n ≠ 0 ⇒ n ≠ 9Also, k can't be zero because the equation is quadratic, so the coefficient of x² is k. So k ≠ 0. But in our expressions for k, 2/(6 - m) and 1/(9 - n), if m and n are such that denominators are non-zero, then k is defined. So our main goal is to find integers m and n such that m = 2n - 12 and k = 2/(6 - m) = 1/(9 - n). Let's substitute m from the first equation into the expression for k.Since m = 2n - 12, then:k = 2/(6 - (2n - 12)) = 2/(6 - 2n + 12) = 2/(18 - 2n) = 2/(2(9 - n)) ) = 1/(9 - n)Which matches the second equation, so that's consistent.Therefore, n can be any integer except 9 (since 9 - n ≠ 0), and m is determined by m = 2n - 12. Then, k is given by k = 1/(9 - n). So k is determined by the integer n ≠ 9. However, we also need to ensure that the original quadratic equation has two real roots. Wait, but the problem says "find all real numbers k such that the quadratic equation... has two integer roots." So even if k is real, but the equation has two integer roots, which must be real. But quadratics can have two real roots or complex roots. Since we need two integer roots, they must be real. So the discriminant must be non-negative, but since we already have that the roots are integers, the discriminant is a perfect square, which is non-negative. So the discriminant condition is automatically satisfied once we have the roots as integers.But maybe there's a case where even though the expressions for p + q and pq are integers, the roots might not be real? Wait, no. If the discriminant is a perfect square, then the roots are real. So if the roots are integers, the discriminant is a perfect square, hence the roots are real. So we don't need to worry about that. So our main constraints are that 6 - 2/k and 9 - 1/k are integers.So let's consider 9 - 1/k must be an integer. Let me denote 9 - 1/k = n ⇒ 1/k = 9 - n ⇒ k = 1/(9 - n), where n is an integer. Similarly, 6 - 2/k must be an integer. Let's express 2/k in terms of n. Since k = 1/(9 - n), 2/k = 2(9 - n). So 6 - 2(9 - n) must be an integer. Let's compute that:6 - 2(9 - n) = 6 - 18 + 2n = 2n - 12. Which is equal to m, which is an integer as per earlier. So this is consistent. So n is any integer except 9, which would make k undefined. So k = 1/(9 - n) where n is an integer. But we also need that k is such that the quadratic equation has two integer roots. Wait, but how do we ensure that?Because Vieta's formulas give us the sum and product as integers, but in general, even if the sum and product are integers, the roots might not be integers. For example, if sum is 3 and product is 2, roots are 1 and 2, which are integers. But if sum is 5 and product is 6, roots are 2 and 3. But if sum is 1 and product is 2, roots are not integers. So Vieta's gives us necessary conditions but not sufficient. Therefore, even if we have sum and product as integers, the roots might not be integers. So we need a stronger condition.Therefore, maybe my initial approach was incomplete. Let me think again.Given that the quadratic equation has two integer roots, let's denote them as p and q. Then the quadratic can be factored as k(x - p)(x - q) = 0. Expanding this, we get kx² - k(p + q)x + kpq = 0. Comparing coefficients with the given equation:Given equation: kx² - 2(3k - 1)x + (9k - 1) = 0So coefficients:- Leading coefficient: k = k (so that's consistent)- Linear coefficient: -k(p + q) = -2(3k - 1) ⇒ k(p + q) = 2(3k - 1)- Constant term: kpq = 9k - 1So we have two equations:1. k(p + q) = 6k - 2 ⇒ k(p + q - 6) = -2 ⇒ k = -2/(p + q - 6)2. kpq = 9k - 1 ⇒ k(pq - 9) = -1 ⇒ k = -1/(pq - 9)Therefore, equating the two expressions for k:-2/(p + q - 6) = -1/(pq - 9) ⇒ 2/(p + q - 6) = 1/(pq - 9) ⇒ 2(pq - 9) = p + q - 6 ⇒ 2pq - 18 = p + q - 6 ⇒ 2pq - p - q = 12So we have the equation 2pq - p - q = 12. Let's see if we can solve this for integers p and q.Hmm, this seems a bit tricky. Let's rearrange the equation:2pq - p - q = 12 ⇒ (2pq - p - q) = 12 ⇒ Let's factor:p(2q - 1) - q = 12 ⇒ p(2q - 1) - q = 12Alternatively, maybe add 1 to both sides to factor:2pq - p - q + 1 = 13 ⇒ (2p - 1)(2q - 1) = 13. Wait, let's check that.Wait, (2p - 1)(2q - 1) = 4pq - 2p - 2q + 1. But in our case, 2pq - p - q = 12. So 4pq - 2p - 2q = 24. Then, 4pq - 2p - 2q + 1 = 25 ⇒ (2p - 1)(2q - 1) = 25. Wait, that seems promising.Let me verify:Starting from 2pq - p - q = 12.Multiply both sides by 2: 4pq - 2p - 2q = 24Add 1 to both sides: 4pq - 2p - 2q + 1 = 25Left-hand side factors as (2p - 1)(2q - 1). Let's check:(2p - 1)(2q - 1) = 4pq - 2p - 2q + 1. Exactly. So:(2p - 1)(2q - 1) = 25So now, since p and q are integers, 2p - 1 and 2q - 1 must be integers that multiply to 25. So the pairs of integers (a, b) such that ab = 25 are:(1, 25), (25, 1), (-1, -25), (-25, -1), (5, 5), (-5, -5)Therefore, we can set 2p - 1 = a and 2q - 1 = b, where (a, b) is any of these factor pairs. Then solve for p and q.Let's go through each factor pair:1. (1, 25):2p - 1 = 1 ⇒ 2p = 2 ⇒ p = 12q - 1 = 25 ⇒ 2q = 26 ⇒ q = 13So (p, q) = (1, 13)2. (25, 1):2p - 1 = 25 ⇒ p = 132q - 1 = 1 ⇒ q = 1So (13, 1) which is the same as the previous one.3. (-1, -25):2p - 1 = -1 ⇒ p = 02q - 1 = -25 ⇒ q = -12But wait, p = 0 would be a root, so substituting back into the quadratic equation, when x=0, the equation becomes 0 - 0 + 9k -1 =0 ⇒ 9k -1 =0 ⇒ k = 1/9. But let's check if p=0 and q=-12 are valid.But if the roots are 0 and -12, then Vieta's formula gives sum = 0 + (-12) = -12, which should equal 2(3k -1)/k. Let's compute:If k=1/9, then 2(3*(1/9) -1)/(1/9) = 2*(1/3 -1)/(1/9) = 2*(-2/3)/(1/9) = 2*(-2/3)*9 = 2*(-6) = -12. So that's correct. Similarly, product is 0*(-12) = 0, which should equal (9k -1)/k. If k=1/9, then (9*(1/9) -1)/ (1/9) = (1 -1)/ (1/9) = 0/(1/9) =0. So that's correct. So even though 0 is a root, it's acceptable because the problem states "two integer roots", and 0 is an integer. So that's valid.4. (-25, -1):2p -1 = -25 ⇒ p = -122q -1 = -1 ⇒ q = 0Same as above, (p, q)= (-12,0)5. (5,5):2p -1 =5 ⇒ p=32q -1=5 ⇒ q=3So (3,3). A double root at 3.6. (-5, -5):2p -1 =-5 ⇒ p=-22q -1=-5 ⇒ q=-2Double root at -2.So these are all possible factor pairs. Therefore, the possible integer roots are:(1,13), (13,1), (0,-12), (-12,0), (3,3), (-2,-2)Now, for each of these pairs, we can compute k using one of the expressions we had earlier. Let's use k = -2/(p + q -6) or k = -1/(pq - 9). Let's compute both to verify.First pair: (1,13)Sum p + q = 14Product pq = 13Compute k using sum:k = -2/(14 -6) = -2/8 = -1/4Check using product:k = -1/(13 -9) = -1/4. Correct.Second pair: (13,1) same as above.Third pair: (0,-12)Sum p + q = -12Product pq = 0Compute k using sum:k = -2/(-12 -6) = -2/(-18) = 1/9Check using product:k = -1/(0 -9) = -1/(-9) = 1/9. Correct.Fourth pair: (-12,0) same as above.Fifth pair: (3,3)Sum p + q =6Product pq=9k using sum:k = -2/(6 -6) = -2/0. Uh-oh, division by zero. That's undefined. Wait, but using product:k = -1/(9 -9) = -1/0, also undefined. So what's going on here?But (3,3) was a solution from the factorization. Wait, but in our original step when we had (2p -1)(2q -1)=25, when (5,5), that leads to p=3 and q=3. But plugging into the equations for k, we get division by zero. That's inconsistent. So does that mean that (3,3) is not a valid solution?Wait, let's check the original quadratic equation. If the roots are both 3, then the quadratic equation would be k(x -3)^2 =0 ⇒ kx² -6kx +9k =0. Comparing with the given equation:Given: kx² -2(3k -1)x +9k -1=0So coefficients:- Leading: k vs k (same)- Linear term: -6k vs -2(3k -1) = -6k +2- Constant term:9k vs 9k -1Therefore, to have the quadratic equation with double root at 3, we need:-6k = -6k +2 ⇒ 0=2, which is impossible. Similarly, constant term:9k =9k -1 ⇒ 0=-1, impossible. Therefore, there is no such k that gives a double root at 3. Hence, the pair (3,3) doesn't lead to a valid k. Therefore, even though (3,3) solves the equation (2p -1)(2q -1)=25, when we substitute back, it leads to a division by zero in k, which is undefined. Therefore, (3,3) is not a valid solution. Similarly, check (-2,-2):Sum p + q = -4Product pq =4Compute k using sum:k = -2/(-4 -6) = -2/(-10)=1/5Check using product:k = -1/(4 -9)= -1/(-5)=1/5. Correct.But wait, let's check if (-2,-2) actually satisfies the quadratic equation. Let's compute k=1/5.Then the quadratic equation becomes (1/5)x² -2(3*(1/5) -1)x +9*(1/5) -1=0Simplify:(1/5)x² -2*(3/5 -1)x +9/5 -1=0= (1/5)x² -2*(-2/5)x +4/5=0= (1/5)x² + (4/5)x +4/5=0Multiply through by 5: x² +4x +4=0 ⇒ (x +2)^2=0. So double root at x=-2. Which is indeed an integer root. Wait, but Vieta's sum is -4 and product is4. But in our calculation, k=1/5. So that's valid. But when we had the pair (3,3), the k is undefined, but for (-2,-2), k=1/5 is valid. So why the difference?Because when we did the factorization (2p -1)(2q -1)=25, for the pair (5,5), leading to p=3, q=3, we get division by zero in k. But for (-5,-5), leading to p=-2, q=-2, we get a valid k. So the difference is that when we have (a,b)=(-5,-5), even though (2p -1)(2q -1)=25, p=-2, q=-2, we can compute k. So maybe (3,3) is invalid, but (-2,-2) is valid. Hence, we need to check each pair.So let's list all possible pairs:From the factor pairs of 25:1. (1,25) ⇒ (p,q)=(1,13), k=-1/42. (25,1) ⇒ (13,1), same as above3. (-1,-25) ⇒ (0,-12), k=1/94. (-25,-1) ⇒ (-12,0), same as above5. (5,5) ⇒ (3,3), invalid (k undefined)6. (-5,-5) ⇒ (-2,-2), k=1/5So the valid pairs are (1,13), (13,1), (0,-12), (-12,0), (-2,-2)So these correspond to k=-1/4, 1/9, and 1/5.Wait, (1,13) and (13,1) give same k=-1/4. Similarly, (0,-12) and (-12,0) give k=1/9. And (-2,-2) gives k=1/5.Therefore, the possible k values are -1/4, 1/9, and 1/5. Let's check each of them.First, k = -1/4:Quadratic equation: (-1/4)x² -2(3*(-1/4) -1)x +9*(-1/4) -1=0Compute each term:Leading: -1/4 x²Linear term: -2*(-3/4 -1) = -2*(-7/4) = 14/4 = 7/2, so +7/2 xConstant term: -9/4 -1 = -13/4Thus equation: (-1/4)x² + (7/2)x -13/4 =0. Multiply through by -4:x² -14x +13=0Factor: (x -1)(x -13)=0. Roots 1 and 13. Correct.Second, k=1/9:Equation: (1/9)x² -2(3*(1/9)-1)x +9*(1/9)-1=0Compute terms:Leading: 1/9 x²Linear term: -2*(1/3 -1)= -2*(-2/3)=4/3 xConstant term:1 -1=0Equation: (1/9)x² + (4/3)x =0. Multiply by 9:x² +12x=0 ⇒ x(x+12)=0. Roots 0 and -12. Correct.Third, k=1/5:Equation: (1/5)x² -2(3*(1/5)-1)x +9*(1/5)-1=0Compute terms:Leading:1/5 x²Linear term: -2*(3/5 -1)= -2*(-2/5)=4/5 xConstant term:9/5 -1=4/5Equation: (1/5)x² + (4/5)x +4/5=0. Multiply by 5:x² +4x +4=0 ⇒ (x +2)^2=0. Double root at -2. Correct.Therefore, all three values of k give quadratics with integer roots. However, the case with double root at -2 is acceptable because it's still two integer roots (they are the same, but still integers). The problem says "two integer roots", which can be two equal integers. So multiplicity doesn't matter as long as the roots are integers.So the possible k's are k=-1/4, 1/9, and 1/5.Wait, but let's check if there are any other factor pairs. Wait, 25 can also be factored as (5,5) and (-5,-5), but we saw that (5,5) leads to invalid k, but (-5,-5) leads to valid k. So I think we covered all possibilities.Therefore, the real numbers k are -1/4, 1/9, and 1/5.Final AnswerThe real numbers ( k ) are (boxed{-dfrac{1}{4}}), (boxed{dfrac{1}{5}}), and (boxed{dfrac{1}{9}}).</think>To find all real numbers ( k ) such that the quadratic equation ( kx^2 - 2(3k - 1)x + 9k - 1 = 0 ) has two integer roots, we start by using Vieta's formulas. If the roots are integers ( p ) and ( q ), then:1. ( p + q = frac{2(3k - 1)}{k} )2. ( pq = frac{9k - 1}{k} )These expressions must be integers. Simplifying, we get:1. ( 6 - frac{2}{k} ) must be an integer.2. ( 9 - frac{1}{k} ) must be an integer.Let ( 6 - frac{2}{k} = m ) and ( 9 - frac{1}{k} = n ), where ( m ) and ( n ) are integers. Solving for ( k ):1. ( k = frac{2}{6 - m} )2. ( k = frac{1}{9 - n} )Equating these expressions, we find ( m = 2n - 12 ). Substituting back, we determine that ( k = frac{1}{9 - n} ).Next, we consider the quadratic equation in terms of its roots ( p ) and ( q ). The equation can be factored as ( k(x - p)(x - q) = 0 ). Expanding and comparing coefficients, we derive the equations:1. ( k(p + q) = 6k - 2 )2. ( kpq = 9k - 1 )Solving these, we find ( 2pq - p - q = 12 ). Rearranging and factoring, we get ( (2p - 1)(2q - 1) = 25 ). The factor pairs of 25 are considered, leading to possible roots:1. ( (1, 13) ) and ( (13, 1) )2. ( (0, -12) ) and ( (-12, 0) )3. ( (-2, -2) )For each pair, we calculate ( k ):1. For ( (1, 13) ) and ( (13, 1) ), ( k = -frac{1}{4} )2. For ( (0, -12) ) and ( (-12, 0) ), ( k = frac{1}{9} )3. For ( (-2, -2) ), ( k = frac{1}{5} )Verifying each ( k ) confirms the quadratic equation has integer roots. Thus, the real numbers ( k ) are:[boxed{-dfrac{1}{4}}, boxed{dfrac{1}{5}}, text{ and } boxed{dfrac{1}{9}}]]

🔑:Let the equation bek x^{2}-2(3 k-1) x+9 k-1=0with two integer roots x_{1}, x_{2}, and x_{1} leqslant x_{2}. Thenx_{1}+x_{2}=6-frac{2}{k}, x_{1} x_{2}=9-frac{1}{k} text {. }Eliminating frac{1}{k} from the above two equations, we get2 x_{1} x_{2}-x_{1}-x_{2}=12 text {, }which simplifies to left(2 x_{1}-1right)left(2 x_{2}-1right)=5^{2}, and 2 x_{1}-1 leqslant 2 x_{2}-1.left{begin{array} { l } { 2 x _ { 1 } - 1 = - 2 5 , } { 2 x _ { 2 } - 1 = - 1 ; }end{array} left{begin{array}{l}2 x_{1}-1=-5, 2 x_{2}-1=-5 .end{array}right.right.Solving these, we get left{begin{array}{l}x_{1}=1, x_{2}=13 ;end{array}left{begin{array}{l}x_{1}=3 x_{2}=3 ;end{array}left{begin{array}{l}x=-12, x_{2}=0 ;end{array}left{begin{array}{l}x_{1}=-2, x_{2}=-2 .end{array}right.right.right.right.Upon verification, x_{1}=x_{2}=3 is an extraneous root, hencek=frac{1}{9-x_{1} x_{2}}=-frac{1}{4}, frac{1}{9}, frac{1}{5} .

🧠:[<think>Okay, let's try to tackle this problem step by step. The question is about rotating the graph of a function 90 degrees counterclockwise around the point (1, -1) and determining the equation of the resulting curve. The answer choices involve the inverse function, so I need to recall how transformations and rotations affect functions and their inverses.First, let's remember that rotating a graph 90 degrees counterclockwise generally involves a transformation. If we have a point (a, b) on the original graph, after a 90-degree counterclockwise rotation around the origin, the new coordinates would be (-b, a). However, in this case, the rotation is around the point (1, -1), not the origin. So, I need to adjust the coordinates so that the center of rotation is effectively the origin, perform the rotation, and then shift back.Let me break it down:1. Translate the graph so that the center of rotation (1, -1) becomes the origin. To do this, we subtract the coordinates of the center from each point on the graph. So, if a point on the original graph is (x, y), the translated point would be (x - 1, y + 1). Let's denote the translated coordinates as (x', y'), where x' = x - 1 and y' = y + 1. In terms of the original function, y = f(x), so the translated function becomes y' + (-1) = f(x' + 1). Wait, actually, maybe it's better to express y in terms of x. Let's see. Original coordinates: (x, y) where y = f(x). After translating, the coordinates are (x', y') = (x - 1, y + 1). So, x = x' + 1 and y = y' - 1. Therefore, substituting into the original equation: y' - 1 = f(x' + 1). Hence, the translated function is y' = f(x' + 1) + 1. Wait, no, let me check again. If (x, y) becomes (x' = x - 1, y' = y + 1), then the original y = f(x) becomes (y' - 1) = f(x' + 1). Therefore, y' = f(x' + 1) + 1. Hmm, actually, no. Let me think. Wait, if we have a point (x, y) on the original graph, after translating, the new coordinates are (x - 1, y + 1). So, the original point (x, y) is equivalent to the translated point (x', y') where x = x' + 1 and y = y' - 1. Since the original equation is y = f(x), substituting gives y' - 1 = f(x' + 1). Therefore, the translated equation is y' = f(x' + 1) + 1. Hmm, actually, that seems correct. So, the translated function in terms of x' and y' is y' = f(x' + 1) + 1.2. Apply the 90-degree counterclockwise rotation around the new origin (which was the original center (1, -1)). As I mentioned earlier, a 90-degree counterclockwise rotation around the origin transforms a point (a, b) to (-b, a). So, applying this to the translated coordinates (x', y'), the rotated coordinates (x'', y'') would be x'' = -y', y'' = x'. Let me verify that. If we have a point (x', y') and rotate it 90 degrees counterclockwise around the origin, yes, the new coordinates become (-y', x'). So, x'' = -y', y'' = x'.3. Express the rotated coordinates in terms of the original coordinates. Wait, actually, after rotating, we need to translate back to the original coordinate system. But hold on, no. Wait, we first translated the graph so that (1, -1) becomes the origin, then rotated it, and now we need to translate back. Wait, no. Let me think again. The process should be: Translate the graph so that (1, -1) becomes the origin. Then rotate the translated graph 90 degrees counterclockwise. Then translate back by adding (1, -1) to the rotated coordinates. Wait, actually, the sequence is: - Original coordinates (x, y). - Translate to center: (x - 1, y + 1). - Rotate 90 CCW: (- (y + 1), x - 1). Because the rotation of (a, b) around origin is (-b, a). So here, a = x - 1, b = y + 1. So rotated coordinates would be (- (y + 1), x - 1). - Then translate back by adding the center (1, -1). Wait, no. Wait, when you rotate around the center, the rotation is applied after translating the center to the origin, and then translating back. So the steps are: 1. Take a point (x, y) on the original graph. 2. Translate it by subtracting the center: (x - 1, y + 1). 3. Apply rotation: (- (y + 1), x - 1). 4. Translate back by adding the center: ( - (y + 1) + 1, (x - 1) + (-1) ) = ( -y - 1 + 1, x - 1 - 1 ) = ( -y, x - 2 ). So, the rotated point (x_rotated, y_rotated) is ( -y, x - 2 ). Therefore, for every point (x, y) on the original graph, the rotated point is ( -y, x - 2 ). So, to find the equation of the rotated graph, we need to express x_rotated and y_rotated in terms of x and y, and then eliminate x and y using the original equation y = f(x). Let me write that down. Let (X, Y) be the coordinates after rotation. So, X = -y Y = x - 2 We need to express Y in terms of X. Since the original equation is y = f(x), we can express x and y in terms of X and Y. From the first equation, X = -y ⇒ y = -X. From the second equation, Y = x - 2 ⇒ x = Y + 2. Since y = f(x), substituting y and x gives: -X = f(Y + 2) Now, solve for Y: Y + 2 = f^{-1}(-X) ⇒ Y = f^{-1}(-X) - 2 Therefore, the equation of the rotated graph is Y = f^{-1}(-X) - 2. But since X and Y are just the coordinates after rotation, we can replace them with x and y for the final equation: y = f^{-1}(-x) - 2 Comparing this with the answer choices: (A) y = f^{-1}(-x) - 2 (B) y = -f^{-1}(-x) - 2 (C) y = f^{-1}(-x +1) -1 (D) y = f^{-1}(-x -1) +1 So, the correct answer should be (A). But wait, the options are given as (A) through (D), and (A) is exactly what we derived: y = f^{-1}(-x) - 2. However, let me double-check my steps to make sure I didn't make a mistake. Let me go through the steps again. 1. Original point: (x, y) where y = f(x). 2. Translate to center (1, -1): x' = x - 1 y' = y + 1 3. Rotate 90 degrees CCW around origin: x'' = -y' = - (y + 1) y'' = x' = x - 1 4. Translate back: X = x'' + 1 = - (y + 1) + 1 = -y -1 +1 = -y Y = y'' + (-1) = (x - 1) -1 = x - 2 So, the rotated coordinates (X, Y) are (-y, x - 2). Therefore, X = -y and Y = x - 2. So, solving for x and y: x = Y + 2 y = -X Since y = f(x), substituting gives: -X = f(Y + 2) ⇒ Y + 2 = f^{-1}(-X) ⇒ Y = f^{-1}(-X) - 2 So, Y = f^{-1}(-X) - 2, which in terms of the rotated coordinates (X, Y) is the equation. Therefore, replacing X and Y with x and y (since after all, the rotated graph is in the same coordinate system), we get: y = f^{-1}(-x) - 2 Which is answer choice (A). However, let me check if any of the other answer choices could be equivalent or if I made a mistake in the transformation. Wait, could there be a reflection or inversion I missed? Let's consider an example. Suppose f(x) is a simple function, say f(x) = x. Then f^{-1}(x) = x. Rotating the line y = x 90 degrees counterclockwise around (1, -1). Let's see what the answer would be. If f(x) = x, then according to answer (A), the rotated equation is y = f^{-1}(-x) - 2 = (-x) - 2. So, y = -x - 2. Let's see if that's correct. Let's take a point on y = x, say (1,1). Translate it by (-1, +1): (0, 2). Rotate 90 CCW: (-2, 0). Translate back by (1, -1): (-2 +1, 0 -1) = (-1, -1). So, the rotated point is (-1, -1). Let's check if this lies on y = -x - 2. Plug in x = -1: y = -(-1) -2 = 1 -2 = -1. Correct. Another point: (2,2). Translate: (1,3). Rotate: (-3,1). Translate back: (-3 +1,1 -1) = (-2, 0). On y = -x -2: y = -(-2) -2 = 2 -2 = 0. Correct. So, it seems answer (A) works for this case. Let me check answer (B): y = -f^{-1}(-x) -2. In this case, with f^{-1}(x) = x, this would be y = -(-x) -2 = x -2. But for the point (-1, -1), plugging in x = -1: y = (-1) -2 = -3 ≠ -1. So (B) is incorrect here. So, (A) seems correct in this example. Therefore, maybe (A) is the right answer. But let me check another example where f is invertible but not linear. Let's take f(x) = 2x + 3. Then f^{-1}(x) = (x - 3)/2. The original graph is a line. Rotating it 90 CCW around (1, -1). Let's compute the equation using the method above and see if it matches (A). According to (A), the rotated equation should be y = f^{-1}(-x) -2 = [(-x -3)/2] -2 = (-x -3)/2 - 2 = (-x -3 -4)/2 = (-x -7)/2. Let's do the rotation manually. Take a point on f(x) = 2x + 3. Let's take x = 1, then y = 5. Translate to center: (0,6). Rotate 90 CCW: (-6,0). Translate back: (-6 +1, 0 -1) = (-5, -1). Now, check if (-5, -1) lies on y = (-x -7)/2. Plug in x = -5: y = (-(-5) -7)/2 = (5 -7)/2 = (-2)/2 = -1. Correct. Another point: x = 0, y = 3. Translate: (-1,4). Rotate: (-4, -1). Translate back: (-4 +1, -1 -1) = (-3, -2). On the equation y = (-x -7)/2: x = -3, y = (-(-3) -7)/2 = (3 -7)/2 = (-4)/2 = -2. Correct. So again, (A) works here. Let me check why the other options might not work. For example, option (B) would be y = -f^{-1}(-x) -2. In the first example where f(x)=x, this would give y = -(-x) -2 = x -2, which as we saw doesn't fit the rotated points. Similarly, in the second example, it would be y = -[(-x -3)/2] -2 = (x +3)/2 -2 = (x +3 -4)/2 = (x -1)/2, which is different from the correct equation. So (B) is not correct. Option (C): y = f^{-1}(-x +1) -1. In the first example with f(x)=x, this would be y = (-x +1) -1 = -x. Let's check if that's correct. For the original point (1,1), rotated to (-1, -1). Plugging x = -1 into y = -x gives y = 1, which doesn't match (-1, -1). So (C) is incorrect. Option (D): y = f^{-1}(-x -1) +1. For the first example, this would be y = (-x -1) +1 = -x. Similar problem as (C). For x = -1, y = 1 instead of -1. So incorrect. Therefore, the correct answer should be (A). But wait, the original answer I thought was (A), but let me check again the rotation steps. Wait, in our derivation, we ended up with y = f^{-1}(-x) -2, which is (A). However, looking back at the answer choices, (A) is indeed that. So, unless there's a mistake in the reasoning. Wait, but I recall that rotating the graph of a function 90 degrees counterclockwise generally results in the inverse function with some transformations. For instance, rotating y = f(x) 90 degrees CCW around the origin would result in x = -f^{-1}(-y), but here it's around (1, -1). So, perhaps that's another way to see it. Let me try another approach. Let's recall that rotating a function 90 degrees counterclockwise about a point is equivalent to reflecting it over a line, but perhaps it's easier to stick to coordinate transformations. Alternatively, considering that the inverse function is a reflection over the line y = x. But rotation is different. However, when you rotate a function 90 degrees, the resulting graph might not be a function anymore unless the original function is one-to-one. But since the problem states that f has an inverse function, it must be one-to-one, so the rotated graph should also be a function. But in any case, our earlier derivation seems solid with examples confirming the answer. Therefore, I think the correct answer is (A). Wait, but hold on. Let me verify with another example where the function is not linear, maybe f(x) = x^3. Then f^{-1}(x) = x^{1/3}. Rotating this around (1, -1). Take a point (2, 8) on f(x) = x^3. Translate to center: (1,9). Rotate 90 CCW: (-9,1). Translate back: (-9 +1, 1 -1) = (-8, 0). According to (A), the equation is y = f^{-1}(-x) -2. For x = -8, y = f^{-1}(8) -2 = 2 -2 = 0. Correct. Another point: (0,0) on f(x). Translate: (-1,1). Rotate: (-1, -1). Translate back: (-1 +1, -1 -1) = (0, -2). According to (A): y = f^{-1}(0) -2 = 0 -2 = -2. Correct. So again, (A) works. So, seems like (A) is correct. So why does the answer involve the inverse function? Because rotating 90 degrees counterclockwise essentially swaps x and y with some sign changes, leading to the inverse function. The translation steps adjust for the center of rotation. Therefore, I think the answer is (A).Final Answerboxed{A}</think>Given the function ( y = f(x) ) which has an inverse function ( y = f^{-1}(x) ), we need to determine the equation of the resulting curve after rotating the graph of ( y = f(x) ) 90 degrees counterclockwise around the point ( (1, -1) ).1. Translate the graph to move the center of rotation to the origin: - Original coordinates: ( (x, y) ) - Translated coordinates: ( (x' = x - 1, y' = y + 1) ) - The equation becomes ( y' - 1 = f(x' + 1) ), so ( y' = f(x' + 1) + 1 ).2. Apply a 90-degree counterclockwise rotation around the origin: - The rotation transforms coordinates ( (a, b) ) to ( (-b, a) ). - Translated coordinates ( (x', y') ) become ( (-y', x') ).3. Translate back to the original coordinate system: - Rotated coordinates: ( (-y', x') ) - Translate back: ( (X = -y' + 1, Y = x' - 1) )4. Relate the rotated coordinates to the original function: - For a point ( (x, y) ) on the original graph, the rotated coordinates are ( (X = -y, Y = x - 2) ). - Express ( x ) and ( y ) in terms of ( X ) and ( Y ): - ( X = -y ) implies ( y = -X ) - ( Y = x - 2 ) implies ( x = Y + 2 ) - Substitute into the original function ( y = f(x) ): - ( -X = f(Y + 2) ) - Solve for ( Y ): ( Y + 2 = f^{-1}(-X) ) - Therefore, ( Y = f^{-1}(-X) - 2 )The resulting equation after rotation is ( y = f^{-1}(-x) - 2 ), which corresponds to option (A).[boxed{A}],<think>Okay, so I need to figure out which equation represents the graph of y = f(x) after it's rotated 90 degrees counterclockwise around the point (1, -1). The options are given from A to D. Let me start by recalling how rotations affect coordinates and how inverse functions relate to original functions.First, rotating a graph 90 degrees counterclockwise around a point involves two main steps: translating the coordinate system so that the center of rotation becomes the origin, performing the rotation, and then translating back. Also, I remember that rotating a graph 90 degrees counterclockwise is equivalent to transforming each point (x, y) to (-y, x) if the rotation is around the origin. But since the rotation here is around (1, -1), I need to adjust the coordinates accordingly.Let me break it down step by step.1. Translate the graph so that the center of rotation (1, -1) becomes the origin. To do this, I subtract 1 from all x-coordinates and add 1 to all y-coordinates. So, if a point on the original graph is (a, b), after translation, it becomes (a - 1, b + 1).2. Apply the 90-degree counterclockwise rotation around the new origin. As I mentioned earlier, rotating a point (x, y) 90 degrees counterclockwise around the origin transforms it to (-y, x). So, the translated point (a - 1, b + 1) becomes (-(b + 1), a - 1).3. Translate the rotated coordinates back to the original system. This means adding 1 to the x-coordinate and subtracting 1 from the y-coordinate to reverse the initial translation. So, the rotated point (-(b + 1), a - 1) becomes (-(b + 1) + 1, (a - 1) - 1) = (-b -1 +1, a -1 -1) = (-b, a - 2).So, after the entire rotation process, a point (a, b) on the original graph y = f(x) is transformed to (-b, a - 2). Let me write that as:If (a, b) is on y = f(x), then after rotation, the point becomes (-b, a - 2).Now, let's denote the coordinates after rotation as (X, Y). So,X = -bY = a - 2But since (a, b) lies on y = f(x), we know that b = f(a). So substituting b = f(a) into the equations:X = -f(a)Y = a - 2Now, we need to express Y in terms of X. To do that, we can solve for a from the first equation and substitute into the second equation.From X = -f(a), we can write f(a) = -X. To solve for a, we can apply the inverse function f⁻¹ to both sides:a = f⁻¹(-X)Now substitute this into Y = a - 2:Y = f⁻¹(-X) - 2Therefore, the equation of the rotated graph is Y = f⁻¹(-X) - 2. If we replace X and Y with the standard x and y variables (since the equation is in terms of the new coordinates), we get:y = f⁻¹(-x) - 2Looking at the options given, option A is "y = f⁻¹(-x) - 2", which matches this result. However, let me double-check my steps to make sure I didn't make a mistake.First, translation: subtracting (1, -1) from the original coordinates. Then rotation: 90 degrees counterclockwise. Then translating back by adding (1, -1). Wait, maybe I messed up the direction of translation?Wait, the standard process is:1. Translate the object so that the center of rotation moves to the origin. That means, for a center (h, k), you do (x - h, y - k). In this case, h = 1, k = -1. So, translating the original point (a, b) would be (a - 1, b - (-1)) = (a - 1, b + 1). That's correct.2. Then rotate 90 degrees counterclockwise around the origin: (x, y) becomes (-y, x). So, the translated point (a - 1, b + 1) becomes (- (b + 1), a - 1). Correct.3. Then translate back by adding (h, k), which is (1, -1). So, the rotated point (- (b + 1), a - 1) becomes (- (b + 1) + 1, (a - 1) + (-1)) = (-b -1 +1, a -1 -1) = (-b, a - 2). So that's correct.Then, original point (a, b) on y = f(x) gives (X, Y) = (-b, a - 2). So, solving for Y in terms of X:X = -b => b = -XSince b = f(a), then f(a) = -X => a = f⁻¹(-X)Then Y = a - 2 = f⁻¹(-X) - 2. So Y = f⁻¹(-X) - 2. Therefore, the equation is y = f⁻¹(-x) - 2, which is option A.Wait, but hold on, why is option B also there? Let me check again.Wait, maybe I made a mistake in the rotation direction. When you rotate 90 degrees counterclockwise, the rotation matrix is:[begin{pmatrix}0 & -1 1 & 0 end{pmatrix}]So, for a point (x, y), after translating to origin, rotating, then translating back.Wait, but maybe I confused clockwise and counterclockwise?Wait, no. The standard rotation matrix for 90 degrees counterclockwise is indeed (x, y) -> (-y, x). If it's clockwise, it would be (x, y) -> (y, -x). So, if I rotate counterclockwise, (x, y) becomes (-y, x). That's correct.Wait, but let me visualize. Suppose you have a point (1, 0) on the x-axis. Rotating 90 degrees counterclockwise around the origin would bring it to (0, 1), which is correct. So, yes, (1, 0) becomes (0, 1), so the transformation is (x, y) -> (-y, x). So, that's right.Therefore, the steps seem correct. So, the equation is y = f⁻¹(-x) - 2, which is option A.But wait, let me check the answer choices again:(A) y = f⁻¹(-x) - 2(B) y = -f⁻¹(-x) - 2(C) y = f⁻¹(-x +1) -1(D) y = f⁻¹(-x -1) +1Hmm. So according to my calculation, the answer should be A. But let me check if there's an alternative approach.Another way to think about rotation and inverse functions: the inverse function is the reflection over the line y = x. A rotation might involve a combination of reflections and inverses. However, since the rotation is around a point other than the origin, it complicates things.Alternatively, perhaps using parametric equations. Let's consider parametrizing the original function as x = t, y = f(t). Then, we can apply the rotation to each point (t, f(t)).First, translate the point (t, f(t)) by (-1, 1) to center at (1, -1):Translated point: (t - 1, f(t) + 1)Then apply 90-degree counterclockwise rotation:Rotated point: (- (f(t) + 1), t - 1)Then translate back by (1, -1):Final point: (- (f(t) + 1) + 1, t - 1 - 1) = (-f(t) -1 +1, t - 2) = (-f(t), t - 2)So, the parametric equations for the rotated graph are:x = -f(t)y = t - 2To eliminate the parameter t, solve for t from the second equation:t = y + 2Substitute into the first equation:x = -f(y + 2)But we need to express y in terms of x. Let's solve for y.From x = -f(y + 2), we can rearrange:f(y + 2) = -xApply the inverse function f⁻¹ to both sides:y + 2 = f⁻¹(-x)Therefore, y = f⁻¹(-x) - 2Which is exactly the same result as before. So, this confirms that the equation is y = f⁻¹(-x) - 2, which is option A.Wait, so why is option B even there? Maybe a common mistake is to rotate in the wrong direction? Let me check.If instead, by mistake, someone rotated 90 degrees clockwise instead of counterclockwise, the rotation matrix would be (x, y) -> (y, -x). Let's see what that would lead to.Following the same steps but with clockwise rotation:Translated point (t - 1, f(t) + 1) rotated clockwise becomes (f(t) + 1, - (t - 1)) = (f(t) +1, -t +1)Translate back: (f(t) +1 +1, -t +1 -1) = (f(t) +2, -t)Parametric equations:x = f(t) + 2y = -tSolve for t: t = -ySubstitute into x:x = f(-y) + 2Then solve for y:x - 2 = f(-y)Apply inverse function:f⁻¹(x - 2) = -yThus, y = -f⁻¹(x - 2)Which is different from the options given. So, if someone rotated clockwise, they would get y = -f⁻¹(x - 2), but this is not one of the options. Therefore, the correct rotation direction gives option A.Therefore, the answer should be A.But wait, I need to check the options again. Let me confirm:(A) y = f⁻¹(-x) -2(B) y = -f⁻¹(-x) -2(C) y = f⁻¹(-x +1) -1(D) y = f⁻¹(-x -1) +1Yes, A is the one we derived. Hmm. But let me check with a sample function. Let's take a simple function and see.Suppose f(x) = x. Then f⁻¹(x) = x. The original graph is the line y = x. Rotating this line 90 degrees counterclockwise around (1, -1). What does that look like?Original line y = x. Let's pick a point on the line, say (1,1). Rotating this point 90 degrees counterclockwise around (1, -1). Let's compute where it ends up.First, translate (1,1) by subtracting (1, -1): (1 -1, 1 - (-1)) = (0, 2). Rotate 90 degrees counterclockwise around origin: (-2, 0). Translate back by adding (1, -1): (-2 +1, 0 -1) = (-1, -1). So the point (1,1) on y = x maps to (-1, -1).Another point on y = x: (2,2). Translate: (2 -1, 2 - (-1)) = (1, 3). Rotate: (-3, 1). Translate back: (-3 +1, 1 -1) = (-2, 0). So (2,2) becomes (-2, 0).What does the inverse function look like? Since f⁻¹(x) = x, then option A would be y = f⁻¹(-x) -2 = (-x) -2 = -x -2. Let's see if the points we rotated lie on this line.For the original point (1,1), after rotation, it's (-1, -1). Check if this lies on y = -x -2: plug x=-1, y = -(-1) -2 = 1 -2 = -1. Correct.Another point (-2,0): check y = -x -2. If x=-2, y = -(-2) -2 = 2 -2 = 0. Correct. So, the rotated line y = -x -2 is indeed the correct transformation. So, for f(x) = x, the rotated graph is y = -x -2, which is exactly what option A gives. If we had chosen option B, which is y = -f⁻¹(-x) -2, that would be y = -(-x) -2 = x -2, which is a different line. Testing with the rotated points: (-1, -1) on y = x -2 would mean -1 = -1 -2 = -3, which is false. So option B is incorrect here.Therefore, in this test case, option A works. So, my reasoning seems correct.Alternatively, let's take another function. Let f(x) = 2x + 3. Then f⁻¹(x) = (x - 3)/2.Original function: y = 2x + 3.Let's take a point (0,3) on the original graph. Rotate 90 degrees counterclockwise around (1, -1).Translate (0,3): (0 -1, 3 - (-1)) = (-1, 4). Rotate: (-4, -1). Translate back: (-4 +1, -1 -1) = (-3, -2). So (0,3) becomes (-3, -2).According to option A: y = f⁻¹(-x) -2. Let's compute f⁻¹(-x):f⁻¹(-x) = (-x -3)/2. So y = (-x -3)/2 -2 = (-x -3 -4)/2 = (-x -7)/2.Check if (-3, -2) satisfies this equation. Plug x=-3: y = (-(-3) -7)/2 = (3 -7)/2 = (-4)/2 = -2. Correct.Another point: (1,5) on y=2x+3. Translate (1,5): (0,6). Rotate: (-6,0). Translate back: (-6 +1, 0 -1) = (-5, -1). Check if (-5, -1) lies on y = (-x -7)/2. Plug x=-5: y = (-(-5) -7)/2 = (5 -7)/2 = (-2)/2 = -1. Correct.If we had used option B, which is y = -f⁻¹(-x) -2. Then y = - [(-x -3)/2] -2 = (x +3)/2 -2 = (x +3 -4)/2 = (x -1)/2. For x=-3, y = (-3 -1)/2 = -4/2 = -2. That also gives y=-2. Hmm, but for x=-5, y = (-5 -1)/2 = -6/2 = -3, but the actual y is -1. So in this case, option B would not hold. Wait, but in the first test case, option B didn't work, but here, for x=-3, both options A and B gave the correct y. Hmm, but the second test with x=-5 shows that option B fails. So, the answer must be A.Therefore, I think option A is correct. So why is there a discrepancy when I first thought of the answer? Wait, no discrepancy. My calculations all point to option A. Maybe the test cases further confirm that.But let me check if the answer could be B. Suppose we have a reflection or something. Wait, rotating the graph of y = f(x) 90 degrees counterclockwise is equivalent to inverting the function and then reflecting over some axis?Alternatively, sometimes when you rotate a graph, especially a function, it might not be a function anymore unless it's one-to-one. But since the original function has an inverse, it is one-to-one, so the rotated graph should also be a function.But in any case, according to the steps and test cases, option A seems correct.Wait, but the answer given in some sources might be different. Wait, let me check again the steps.Original point (a, b) becomes after rotation (-b, a - 2). So, if we have (X, Y) = (-b, a - 2). Then, since b = f(a), X = -f(a), and Y = a - 2. So, a = Y + 2. Then, X = -f(Y + 2). Therefore, f(Y + 2) = -X. Then, Y + 2 = f⁻¹(-X), so Y = f⁻¹(-X) - 2. Therefore, Y = f⁻¹(-X) - 2. Which is again option A. So yes, seems correct.Alternatively, maybe there is a mistake in the translation steps. Let me redo the translation steps.Original center is (1, -1). To translate to the origin, we subtract (1, -1). So, for a point (x, y), the translated point is (x - 1, y - (-1)) = (x - 1, y + 1). Then rotate 90 degrees CCW: becomes ( - (y +1), x -1 ). Then translate back by adding (1, -1): ( - (y +1 ) +1, (x -1 ) + (-1) ) = ( -y -1 +1, x -1 -1 ) = ( -y, x - 2 ). Therefore, the new coordinates (X, Y) satisfy X = -y and Y = x - 2. Therefore, solving for x and y in terms of X and Y:From X = -y => y = -XFrom Y = x - 2 => x = Y + 2But since (x, y) is on the original graph, y = f(x). Therefore, -X = f(Y + 2). Therefore, Y + 2 = f⁻¹(-X) => Y = f⁻¹(-X) - 2. Therefore, Y = f⁻¹(-X) - 2. Thus, equation is y = f⁻¹(-x) - 2, which is option A.Therefore, the correct answer is A. But the original problem's options have option B as "y= -f^{-1}(-x)-2". But according to all the steps, there's no negative sign outside the inverse function. So unless there's a miscalculation here.Wait, perhaps I made a mistake in the direction of rotation. Let me check again with the standard rotation.The standard rotation matrices:For 90 degrees counterclockwise: (x, y) → (-y, x)For 90 degrees clockwise: (x, y) → (y, -x)But in our case, after translating, we rotated (x', y') = (x - 1, y + 1) to (-y' , x') = ( - (y +1 ), x -1 ). Then translating back gives ( -y -1 +1, x -1 -1 ) = (-y, x -2 ). So, yes, that's correct.Alternatively, maybe there's a different interpretation. For example, rotating the graph 90 degrees counterclockwise around (1, -1). But maybe the question is rotating the entire coordinate system 90 degrees, which would be equivalent to rotating the graph in the opposite direction. Wait, no. Rotating the graph 90 degrees counterclockwise is the same as rotating the coordinate system 90 degrees clockwise. But in any case, the standard approach is to rotate the points as I did.Alternatively, think of the rotation as a combination of transformations. Since the inverse function is a reflection over y = x, but a rotation involves more. However, all the steps confirm that option A is correct.Therefore, the answer should be A.Final Answerboxed{A}</think>Given the function ( y = f(x) ) has an inverse function ( y = f^{-1}(x) ). We need to determine the equation of the resulting curve after rotating the graph of ( y = f(x) ) 90 degrees counterclockwise around the point ( (1, -1) ).1. Translate the coordinates so that the center of rotation ((1, -1)) becomes the origin: - For a point ((a, b)) on the graph ( y = f(x) ), the translated coordinates are ((a - 1, b + 1)).2. Apply a 90-degree counterclockwise rotation around the origin: - The rotation transforms ((x, y)) to ((-y, x)). Thus, the translated point ((a - 1, b + 1)) becomes ((-(b + 1), a - 1)).3. Translate the coordinates back to the original system: - The rotated coordinates ((-(b + 1), a - 1)) are translated back by adding (1) to the x-coordinate and subtracting (1) from the y-coordinate: ((-(b + 1) + 1, (a - 1) - 1) = (-b, a - 2)).4. Relate the new coordinates to the original function: - The new coordinates after rotation are ((X, Y) = (-b, a - 2)). Since ( b = f(a) ), we have ( X = -f(a) ) and ( Y = a - 2 ). - Solving for ( a ) in terms of ( X ): ( a = f^{-1}(-X) ). - Substituting ( a ) into the equation for ( Y ): ( Y = f^{-1}(-X) - 2 ).Thus, the equation of the resulting curve is ( y = f^{-1}(-x) - 2 ), which corresponds to option (boxed{A}).]

🔑:2. A.The point (t, f(t)) is rotated 90^{circ} counterclockwise around (1, -1), resulting in (-f(t), t-2).Let x = -f(t). Theny = t-2 = f^{-1}(-x) - 2 text{.}

🧠:[<think>Okay, let's try to solve this arithmetic sequence problem. Hmm, part (I) asks for the general term formula and the sum of the first n terms. I remember that in an arithmetic sequence, each term is the previous term plus a common difference. So, the general term is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. They give us that a_3 = 7. Let me write that down. Since a_3 is the third term, substituting n=3 into the general formula: a_3 = a_1 + 2d = 7. That's one equation. They also say a_5 + a_7 = 26. Let's write expressions for a_5 and a_7. a_5 would be a_1 + 4d, and a_7 is a_1 + 6d. Adding them together: (a_1 + 4d) + (a_1 + 6d) = 2a_1 + 10d = 26. So now we have two equations:1) a_1 + 2d = 72) 2a_1 + 10d = 26I need to solve these two equations for a_1 and d. Let's maybe solve the first equation for a_1: a_1 = 7 - 2d. Then substitute that into the second equation. So, substituting into equation 2: 2*(7 - 2d) + 10d = 26. Let's compute that:14 - 4d + 10d = 26Combine like terms: 14 + 6d = 26Subtract 14 from both sides: 6d = 12Divide by 6: d = 2Then substitute d = 2 back into a_1 = 7 - 2d: a_1 = 7 - 4 = 3So, the first term is 3 and the common difference is 2. Therefore, the general term formula is a_n = 3 + (n - 1)*2. Let me simplify that:a_n = 3 + 2n - 2 = 2n + 1. Wait, 3 - 2 is 1, so a_n = 2n + 1. Wait, hold on, 3 + (n-1)*2 = 3 + 2n - 2 = 2n + 1? No, wait, 3 - 2 is 1, so it's 2n +1. Let me check with n=3: a_3 should be 7. 2*3 +1 =7, yes. And a_5 is 2*5 +1=11, a_7=2*7+1=15, so 11 +15=26. That checks out. So the general term is a_n = 2n +1.Now, the sum of the first n terms, S_n. The formula for the sum of an arithmetic sequence is S_n = n/2*(a_1 + a_n). Since we know a_1 is 3 and a_n is 2n +1, substitute those in:S_n = n/2*(3 + 2n +1) = n/2*(2n +4) = n/2*2(n +2) = n(n +2). So S_n = n(n +2). Let me verify that. For example, if n=3, sum S_3 should be 3 +5 +7=15. Using the formula, 3*(3+2)=3*5=15, correct. Similarly, S_1 should be 3, and 1*(1+2)=3, correct. Good.So part (I) is done. The general term is a_n = 2n +1 and S_n = n(n +2).Now part (II): Let b_n = 1/(S_n - n), find the sum of the first 8 terms of {b_n}.First, let's compute S_n - n. Since S_n = n(n +2), then S_n -n = n(n +2) -n = n(n +2 -1) = n(n +1). Therefore, b_n = 1/(n(n +1)).So, b_n = 1/(n(n +1)). Hmm, this is a telescoping series, right? Because 1/(n(n +1)) can be written as 1/n - 1/(n +1). Let me check:1/(n(n +1)) = (1/n) - (1/(n +1)). Multiply both sides by n(n +1):1 = (n +1) - n = 1. Correct. So that's a telescoping series.Therefore, the sum of the first 8 terms of {b_n} is the sum from n=1 to 8 of [1/n - 1/(n +1)]. When we expand this:(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/8 - 1/9)Each term cancels with the next one: -1/2 cancels with +1/2, -1/3 cancels with +1/3, etc. So all that's left is the first term of the first expression, 1/1, and the last term of the last expression, -1/9. Therefore, the sum is 1 - 1/9 = 8/9.Let me verify this with actual computation. Let's compute each b_n:For n=1: b_1 = 1/(1*2) = 1/2n=2: 1/(2*3)=1/6n=3: 1/12n=4:1/20n=5:1/30n=6:1/42n=7:1/56n=8:1/72Sum them up:1/2 +1/6 = (3/6 +1/6)=4/6=2/32/3 +1/12= (8/12 +1/12)=9/12=3/43/4 +1/20= (15/20 +1/20)=16/20=4/54/5 +1/30= (24/30 +1/30)=25/30=5/65/6 +1/42= (35/42 +1/42)=36/42=6/76/7 +1/56= (48/56 +1/56)=49/56=7/87/8 +1/72= (63/72 +1/72)=64/72=8/9Yes, that's 8/9. So the sum is indeed 8/9.Therefore, part (II) answer is 8/9.Final Answer(I) The general term is boxed{a_n = 2n + 1} and the sum is boxed{S_n = n(n + 2)}; (II) The sum of the first 8 terms of ({b_n}) is boxed{dfrac{8}{9}}.</think>Given an arithmetic sequence ({a_n}) that satisfies (a_3 = 7), (a_5 + a_7 = 26), and its sum of the first (n) terms is (S_n):(I) To find the general term formula for ({a_n}) and (S_n):1. We know that (a_3 = 7), so (a_1 + 2d = 7).2. We also know that (a_5 + a_7 = 26), so ((a_1 + 4d) + (a_1 + 6d) = 26), which simplifies to (2a_1 + 10d = 26).Solving these equations:1. From (a_1 + 2d = 7), we get (a_1 = 7 - 2d).2. Substituting (a_1 = 7 - 2d) into (2a_1 + 10d = 26): [ 2(7 - 2d) + 10d = 26 implies 14 - 4d + 10d = 26 implies 6d = 12 implies d = 2 ]3. Substituting (d = 2) back into (a_1 = 7 - 2d) gives (a_1 = 3).Thus, the general term formula is (a_n = a_1 + (n-1)d = 3 + (n-1) cdot 2 = 2n + 1).For the sum (S_n):[S_n = frac{n}{2} (a_1 + a_n) = frac{n}{2} (3 + (2n + 1)) = frac{n}{2} (2n + 4) = n(n + 2)](II) To find the sum of the first 8 terms of the sequence ({b_n}) where (b_n = frac{1}{S_n - n}):1. Calculate (S_n - n): [ S_n - n = n(n + 2) - n = n(n + 1) ]2. Thus, (b_n = frac{1}{n(n + 1)}), which can be written as a telescoping series: [ b_n = frac{1}{n} - frac{1}{n + 1} ]Summing the first 8 terms:[sum_{k=1}^{8} left( frac{1}{k} - frac{1}{k + 1} right) = left(1 - frac{1}{2}right) + left(frac{1}{2} - frac{1}{3}right) + cdots + left(frac{1}{8} - frac{1}{9}right)]Most terms cancel out, leaving:[1 - frac{1}{9} = frac{8}{9}] Final Answer(I) The general term is (boxed{a_n = 2n + 1}) and the sum is (boxed{S_n = n(n + 2)}); (II) The sum of the first 8 terms of ({b_n}) is (boxed{dfrac{8}{9}}).,<think>Okay, let's try to solve this arithmetic sequence problem. First, part (I) asks for the general term formula and the sum of the first n terms. Hmm. I remember that in an arithmetic sequence, each term is the previous term plus a common difference, right? So the nth term is given by a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. And the sum S_n is n/2 times (2a_1 + (n-1)d) or also n times (a_1 + a_n)/2. Yeah, that sounds right.We are given that a_3 = 7. Let's write that in terms of a_1 and d. Since a_3 is the third term, it's a_1 + 2d = 7. Got that equation. Then we also have a_5 + a_7 = 26. Let's express those terms as well. The fifth term is a_1 + 4d, and the seventh term is a_1 + 6d. So their sum is (a_1 + 4d) + (a_1 + 6d) = 2a_1 + 10d = 26. So now we have two equations:1) a_1 + 2d = 72) 2a_1 + 10d = 26Hmm, so we can solve this system of equations for a_1 and d. Let me do that. Maybe multiply the first equation by 2 to make the coefficients of a_1 the same. Then we have:1) 2a_1 + 4d = 142) 2a_1 + 10d = 26Subtract the first equation from the second: (2a_1 + 10d) - (2a_1 + 4d) = 26 - 14. That simplifies to 6d = 12, so d = 2. Then substitute back into the first equation: a_1 + 2*2 = 7 => a_1 + 4 = 7 => a_1 = 3. So the first term is 3 and the common difference is 2.Therefore, the general term a_n should be a_1 + (n-1)d = 3 + (n-1)*2. Let me simplify that: 3 + 2n - 2 = 2n +1. Wait, no, 3 - 2 is 1, so it's 2n +1? Wait, 3 + 2(n-1). Let me check again. 3 + 2(n-1) = 3 + 2n - 2 = 2n +1. Yeah, that seems right. So a_n = 2n +1. Let me verify with the given terms. a_3 should be 2*3 +1 = 7. Correct. a_5 is 2*5 +1=11, a_7=2*7 +1=15. Then 11 +15=26. Perfect, that matches the given condition.So the general term is a_n = 2n +1. Now the sum S_n. Using the formula S_n = n/2*(2a_1 + (n-1)d). Plugging in a_1=3 and d=2: S_n = n/2*(2*3 + (n-1)*2) = n/2*(6 + 2n - 2) = n/2*(2n +4) = n*(n +2). So S_n = n(n +2). Alternatively, since a_n = 2n +1, the sum can also be calculated as n*(a_1 + a_n)/2 = n*(3 + 2n +1)/2 = n*(2n +4)/2 = n(n +2). Yep, same result. So part (I) is done: a_n = 2n +1 and S_n = n(n +2).Now part (II): Let b_n = 1/(S_n -n). Find the sum of the first 8 terms of {b_n}.First, let's compute S_n -n. Since S_n = n(n +2), then S_n -n = n(n +2) -n = n(n +2 -1) = n(n +1). So S_n -n = n(n +1). Therefore, b_n = 1/(n(n +1)). Hmm, so b_n is 1/(n(n +1)). That's a standard telescoping series term. The sum of 1/(n(n +1)) from n=1 to 8.I remember that 1/(n(n +1)) can be written as 1/n - 1/(n +1). Let me check: 1/n - 1/(n+1) = (n+1 -n)/(n(n +1)) = 1/(n(n +1)). Exactly. So each term b_n = 1/n -1/(n +1). Therefore, the sum from n=1 to 8 of b_n is (1/1 -1/2) + (1/2 -1/3) + (1/3 -1/4) + ... + (1/8 -1/9). All the intermediate terms cancel out, leaving 1 -1/9 = 8/9. Let me verify that step-by-step.First term: 1 -1/2Second term: +1/2 -1/3Third term: +1/3 -1/4Fourth: +1/4 -1/5Fifth: +1/5 -1/6Sixth: +1/6 -1/7Seventh: +1/7 -1/8Eighth: +1/8 -1/9Adding all these up:1 -1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9Everything cancels except the first 1 and the last -1/9. So total sum is 1 -1/9 = 8/9. So the answer is 8/9.Wait, let me make sure I didn't make a mistake in simplifying S_n -n. S_n is n(n +2), so subtract n: n(n +2) -n = n^2 +2n -n =n^2 +n =n(n +1). Yep, that's correct. So b_n is indeed 1/(n(n +1)). Then the sum is telescoping. So the answer for part (II) is 8/9.Final Answer(I) The general term is (boxed{a_n = 2n + 1}) and the sum is (boxed{S_n = n(n + 2)}).(II) The sum of the first 8 terms of ({b_n}) is (boxed{dfrac{8}{9}}).</think>Given an arithmetic sequence ({a_n}) that satisfies (a_3 = 7), (a_5 + a_7 = 26), and its sum of the first (n) terms is (S_n):(I) To find the general term formula for ({a_n}) and (S_n):1. We know (a_3 = 7), which gives the equation (a_1 + 2d = 7).2. We also know (a_5 + a_7 = 26), which gives the equation (2a_1 + 10d = 26).Solving these equations:- Multiply the first equation by 2: (2a_1 + 4d = 14).- Subtract this from the second equation: (6d = 12 implies d = 2).- Substitute (d = 2) back into the first equation: (a_1 + 4 = 7 implies a_1 = 3).Thus, the general term formula is (a_n = a_1 + (n-1)d = 3 + 2(n-1) = 2n + 1).The sum of the first (n) terms (S_n) is:[S_n = frac{n}{2} left[2a_1 + (n-1)dright] = frac{n}{2} left[6 + 2(n-1)right] = frac{n}{2} left[2n + 4right] = n(n + 2)](II) To find the sum of the first 8 terms of the sequence ({b_n}) where (b_n = frac{1}{S_n - n}):1. Calculate (S_n - n):[S_n - n = n(n + 2) - n = n(n + 1)]2. Thus, (b_n = frac{1}{n(n + 1)}).The sum of the first 8 terms of ({b_n}) is a telescoping series:[sum_{k=1}^{8} left( frac{1}{k} - frac{1}{k+1} right) = left(1 - frac{1}{2}right) + left(frac{1}{2} - frac{1}{3}right) + cdots + left(frac{1}{8} - frac{1}{9}right)]All intermediate terms cancel out, leaving:[1 - frac{1}{9} = frac{8}{9}]Final Answers:(I) The general term is (boxed{a_n = 2n + 1}) and the sum is (boxed{S_n = n(n + 2)}).(II) The sum of the first 8 terms of ({b_n}) is (boxed{dfrac{8}{9}}).]

🔑:(I) Let the common difference of the arithmetic sequence {a_{n}} be d. From a_{5}+a_{7}=26, we obtain a_{6}=13. Moreover, a_{6}-a_{3}=3d=6, which implies d=2.Therefore, the general term formula for {a_{n}} is:a_{n}=a_{3}+(n-3)d=7+2(n-3)=2n+1.And the sum of the first n terms is:S_{n}=frac{a_{1}+a_{n}}{2}times n=frac{3+2n+1}{2}times n=n^{2}+2n.(II) From b_{n}=frac{1}{S_{n}-n}, we have:b_{n}=frac{1}{n^{2}+n}=frac{1}{n(n+1)}=frac{1}{n}-frac{1}{n+1}.Let T_{n} denote the sum of the first n terms of the sequence {b_{n}}. Then,T_{8}=left(1-frac{1}{2}right)+left(frac{1}{2}-frac{1}{3}right)+left(frac{1}{3}-frac{1}{4}right)+cdots+left(frac{1}{8}-frac{1}{9}right)=1-frac{1}{9}=boxed{frac{8}{9}}.

🧠:[<think>Okay, let's tackle this problem step by step. First, part (1) asks for the standard equation of the ellipse C. We know the general form is x²/a² + y²/b² = 1, where a > b > 0. The eccentricity is given as √3/2, and the point (-√3, 1/2) lies on the ellipse.Starting with the eccentricity: for an ellipse, eccentricity e = c/a, where c is the distance from the center to each focus, and c² = a² - b². Since e = √3/2, we have c = (√3/2)a. Then, c² = (3/4)a². Therefore, a² - b² = (3/4)a², which simplifies to b² = a² - (3/4)a² = (1/4)a². So, b² = (a²)/4. That gives a relationship between a and b: b = a/2.Now, since the point (-√3, 1/2) is on the ellipse, plugging into the ellipse equation:( (-√3)² ) / a² + ( (1/2)² ) / b² = 1Calculating that:(3)/a² + (1/4)/b² = 1But we know b² = a²/4, so substitute that in:3/a² + (1/4)/(a²/4) = 3/a² + (1/4)*(4/a²) = 3/a² + 1/a² = 4/a² = 1Therefore, 4/a² = 1 => a² = 4 => a = 2 (since a > 0). Then, since b² = a²/4 = 4/4 = 1, so b = 1.Therefore, the standard equation of the ellipse is x²/4 + y²/1 = 1, or x²/4 + y² = 1.So part (1) is done. The answer is x²/4 + y² = 1.Now part (2): Given a line l intersecting the ellipse at points P and Q, midpoint H of PQ, origin O, and OH = 1. Need to find the maximum area of triangle POQ.First, let's recall that the area of triangle POQ can be calculated using the determinant formula: if P = (x1, y1) and Q = (x2, y2), then area = (1/2)|x1y2 - x2y1|.But since H is the midpoint of PQ, H = ((x1 + x2)/2, (y1 + y2)/2). Given that OH = 1, the distance from the origin to H is 1. So sqrt( [(x1 + x2)/2]^2 + [(y1 + y2)/2]^2 ) = 1. Squaring both sides, [(x1 + x2)/2]^2 + [(y1 + y2)/2]^2 = 1.We need to maximize the area (1/2)|x1y2 - x2y1| given that constraint.Alternatively, perhaps we can parameterize the line l. Let's think about this.Another approach: For a line intersecting an ellipse, the midpoint H can be related to the equation of the line. Let me recall that if a line intersects an ellipse, the coordinates of the midpoint can be found using the equation of the ellipse and the line.Suppose the line l has equation y = kx + m. Then, intersecting with the ellipse x²/4 + y² = 1.Substitute y = kx + m into the ellipse equation:x²/4 + (kx + m)^2 = 1Expand: x²/4 + k²x² + 2kmx + m² = 1Combine terms: (1/4 + k²)x² + 2kmx + (m² - 1) = 0Let this quadratic equation in x have roots x1 and x2. Then, the sum of the roots x1 + x2 = - (2km)/(1/4 + k²). Then, the x-coordinate of H is (x1 + x2)/2 = - (km)/(1/4 + k²).Similarly, the y-coordinates of H: y = kx + m, so the average y-coordinate is k*( (x1 + x2)/2 ) + m = k*(- (km)/(1/4 + k²)) + m = m - (k²m)/(1/4 + k²) = m/(1/4 + k²)*(1/4 + k² - k²) = m/(1/4 + k²)*(1/4) = m/(4*(1/4 + k²)) = m/(1 + 4k²)So H has coordinates ( - (km)/(1/4 + k²), m/(1 + 4k²) )Wait, perhaps there's a miscalculation here. Let's check:y1 = kx1 + my2 = kx2 + mAverage y = (y1 + y2)/2 = k*(x1 + x2)/2 + m = k*(- (km)/(1/4 + k²)) + mWhich is: - (k²m)/(1/4 + k²) + m = m*(1 - k²/(1/4 + k²)) = m*( (1/4 + k² - k²)/ (1/4 + k²) ) = m*(1/4)/(1/4 + k²) = m/(4*(1/4 + k²)) = m/(1 + 4k²)Yes, that's correct.Therefore, coordinates of H are:( - (km)/(1/4 + k²), m/(1 + 4k²) )We also have that the distance from O to H is 1. Therefore,[ - (km)/(1/4 + k²) ]² + [ m/(1 + 4k²) ]² = 1Simplify this equation:( k²m² ) / (1/4 + k²)^2 + m² / (1 + 4k²)^2 = 1Notice that denominator terms: (1/4 + k²) is equal to (1 + 4k²)/4. Therefore, 1/4 + k² = (1 + 4k²)/4. Therefore, (1/4 + k²)^2 = (1 + 4k²)^2 / 16Similarly, (1 + 4k²)^2 is as it is.Therefore, substituting:( k²m² ) / [ (1 + 4k²)^2 / 16 ] + m² / (1 + 4k²)^2 = 1Which is equal to:16k²m² / (1 + 4k²)^2 + m² / (1 + 4k²)^2 = 1Combine terms:(16k²m² + m²) / (1 + 4k²)^2 = 1Factor m²:m²(16k² + 1) / (1 + 4k²)^2 = 1Therefore,m² = (1 + 4k²)^2 / (16k² + 1)Thus, m = ± (1 + 4k²)/sqrt(16k² + 1)So that's an expression for m in terms of k. Now, we need to find the area of triangle POQ.Area = (1/2)|x1y2 - x2y1|But since y1 = kx1 + m and y2 = kx2 + m,x1y2 - x2y1 = x1(kx2 + m) - x2(kx1 + m) = kx1x2 + mx1 - kx1x2 - mx2 = m(x1 - x2)Therefore, Area = (1/2)|m(x1 - x2)|But x1 - x2 can be found from the quadratic equation. The quadratic equation in x is (1/4 + k²)x² + 2kmx + (m² - 1) = 0The difference of roots x1 - x2 is sqrt[(x1 + x2)^2 - 4x1x2]But x1 + x2 = -2km/(1/4 + k²), and x1x2 = (m² - 1)/(1/4 + k²)Therefore, (x1 - x2)^2 = (x1 + x2)^2 - 4x1x2 = [4k²m²/(1/4 + k²)^2] - 4*(m² - 1)/(1/4 + k²)= [4k²m² - 4(m² - 1)(1/4 + k²)] / (1/4 + k²)^2Let me compute numerator:4k²m² - 4(m² - 1)(1/4 + k²)= 4k²m² - 4*( (m²)(1/4 + k²) - (1/4 + k²) )= 4k²m² - 4*( (m²/4 + m²k² - 1/4 - k²) )= 4k²m² - (m² + 4m²k² - 1 - 4k² )= 4k²m² - m² - 4k²m² + 1 + 4k²= -m² + 1 + 4k²Therefore, (x1 - x2)^2 = [ -m² + 1 + 4k² ] / (1/4 + k²)^2But we already have m² = (1 + 4k²)^2 / (16k² + 1)Therefore, substitute that into the numerator:- (1 + 4k²)^2 / (16k² + 1) + 1 + 4k²= [ - (1 + 4k²)^2 + (1 + 4k²)(16k² + 1) ] / (16k² + 1)Factor out (1 + 4k²):= (1 + 4k²)[ - (1 + 4k²) + (16k² + 1) ] / (16k² + 1)Simplify inside the brackets:-1 -4k² +16k² +1 = 12k²Therefore,= (1 + 4k²)(12k²) / (16k² + 1)So (x1 - x2)^2 = [ (1 + 4k²)(12k²) / (16k² + 1) ] / (1/4 + k²)^2But note that 1/4 + k² = (1 + 4k²)/4, so (1/4 + k²)^2 = (1 + 4k²)^2 / 16Therefore,(x1 - x2)^2 = [ (1 + 4k²)(12k²) / (16k² + 1) ] / [ (1 + 4k²)^2 / 16 ]= [12k² / (16k² + 1)] * 16 / (1 + 4k²)= (192k²) / [ (16k² + 1)(1 + 4k²) ]Therefore, |x1 - x2| = sqrt(192k² / [ (16k² + 1)(1 + 4k²) ]) = (sqrt(192)k) / sqrt( (16k² +1)(1 +4k²) )Simplify sqrt(192) = 8*sqrt(3), so:|x1 - x2| = 8√3 k / sqrt( (16k² +1)(1 +4k²) )Therefore, the area is (1/2)|m||x1 - x2| = (1/2)*( (1 + 4k²)/sqrt(16k² +1) )*(8√3 k)/sqrt( (16k² +1)(1 +4k²) ) )Simplify:Multiply the terms:(1/2)*(1 +4k²)/sqrt(16k² +1) * 8√3 k / sqrt( (16k² +1)(1 +4k²) )= (1/2)*8√3 k * (1 +4k²) / sqrt(16k² +1) * 1 / sqrt( (16k² +1)(1 +4k²) )= 4√3 k * (1 +4k²) / sqrt(16k² +1) * 1 / sqrt( (16k² +1)(1 +4k²) )= 4√3 k * (1 +4k²) / [ (16k² +1)^{1/2} * (16k² +1)^{1/2} * (1 +4k²)^{1/2} ) ]Wait, the denominator is sqrt( (16k² +1)(1 +4k²) ) = (16k² +1)^{1/2}*(1 +4k²)^{1/2}So the entire denominator is (16k² +1)^{1/2} * (16k² +1)^{1/2}*(1 +4k²)^{1/2} = (16k² +1) * (1 +4k²)^{1/2}Wait, actually:Wait, the first term is (1 +4k²)/sqrt(16k² +1), then multiplied by 1/sqrt( (16k² +1)(1 +4k²) )So combining the denominators:sqrt(16k² +1) * sqrt( (16k² +1)(1 +4k²) ) = (16k² +1)^{1/2} * (16k² +1)^{1/2} * (1 +4k²)^{1/2} = (16k² +1) * (1 +4k²)^{1/2}So numerator is (1 +4k²), denominator is (16k² +1)*(1 +4k²)^{1/2}Therefore, the entire fraction becomes (1 +4k²)^{1/2} / (16k² +1)Therefore, area = 4√3 k * (1 +4k²)^{1/2} / (16k² +1)Therefore, Area = 4√3 k * sqrt(1 +4k²) / (16k² +1)Simplify this expression. Let me set t = k², so t ≥ 0.Then, Area = 4√3 * sqrt(t) * sqrt(1 +4t) / (16t +1)= 4√3 * sqrt( t(1 +4t) ) / (16t +1)Let me compute t(1 +4t) = t +4t², so sqrt(t +4t²) = sqrt(t(1 +4t)).Alternatively, perhaps we can rationalize this expression. Let's let u = 4t, then t = u/4.Then,sqrt(t +4t²) = sqrt( u/4 +4*(u²)/16 ) = sqrt( u/4 + u²/4 ) = (sqrt(u +u²))/2But not sure if this helps. Alternatively, maybe we can write sqrt(t(1 +4t)) as sqrt(t +4t²). Let's see.Alternatively, perhaps make substitution v = 4t +1, but not sure. Alternatively, perhaps use calculus to maximize the function.Let me denote the area as A(k) = 4√3 k sqrt(1 +4k²) / (16k² +1)We need to maximize A(k) with respect to k.Alternatively, since k can be positive or negative, but area is positive, we can consider k ≥0.Let’s set k ≥0, then A(k) = 4√3 k sqrt(1 +4k²) / (16k² +1)To maximize A(k), take derivative of A(k) with respect to k and set to zero.Let’s compute derivative:Let’s denote f(k) = k sqrt(1 +4k²) / (16k² +1)Compute f'(k):First, write f(k) = k (1 +4k²)^{1/2} / (16k² +1)Use quotient rule:f'(k) = [ ( (1 +4k²)^{1/2} + k*(1/2)(1 +4k²)^{-1/2}*(8k) )*(16k² +1) - k(1 +4k²)^{1/2}*(32k) ] / (16k² +1)^2Simplify step by step.First, compute numerator:First term: derivative of numerator:d/dk [k(1 +4k²)^{1/2}] = (1)(1 +4k²)^{1/2} + k*(1/2)(1 +4k²)^{-1/2}*8k= (1 +4k²)^{1/2} + 4k²(1 +4k²)^{-1/2}= (1 +4k² +4k²)/ (1 +4k²)^{1/2} } = (1 +8k²)/ (1 +4k²)^{1/2}Therefore, derivative of numerator is (1 +8k²)/sqrt(1 +4k²)Multiply by denominator (16k² +1):(1 +8k²)(16k² +1)/sqrt(1 +4k²)Then subtract the second term: numerator times derivative of denominator:k(1 +4k²)^{1/2}*32k=32k²(1 +4k²)^{1/2}Therefore, total numerator:[ (1 +8k²)(16k² +1) -32k²(1 +4k²) ] / sqrt(1 +4k² )Let’s compute the expression in the brackets:Expand (1 +8k²)(16k² +1) = 16k² +1 +128k⁴ +8k² = (16k² +8k²) +1 +128k⁴ =24k² +1 +128k⁴Then subtract 32k²(1 +4k²) =32k² +128k⁴Therefore,24k² +1 +128k⁴ -32k² -128k⁴ = (24k² -32k²) +1 + (128k⁴ -128k⁴) = -8k² +1Therefore, numerator of f'(k) is (-8k² +1)/sqrt(1 +4k²)Therefore, f'(k) = [ (-8k² +1)/sqrt(1 +4k²) ] / (16k² +1)^2Set f'(k) =0 => numerator is zero: -8k² +1=0 => k²=1/8 => k=1/(2√2) or k=-1/(2√2). Since we considered k≥0, k=1/(2√2)Check if this is a maximum. Second derivative test or check sign changes.For k slightly less than 1/(2√2), say k=0.3 (approx 1/(2*1.414)=0.3535), so 0.3 is less. Then f'(k) would be positive (since -8*(0.3)^2 +1= -8*0.09 +1= -0.72 +1=0.28 >0. So derivative is positive before k=1/(2√2), and negative after, so it's a maximum.Therefore, maximum at k=1/(2√2)Now compute A(k) at k=1/(2√2):First, compute 4√3 * k * sqrt(1 +4k²)/(16k² +1)Compute k =1/(2√2)Compute 4√3*(1/(2√2)) * sqrt(1 +4*(1/(2√2))²)/(16*(1/(2√2))² +1)First, compute 4√3*(1/(2√2)) = (4√3)/(2√2) = 2√3/√2 = 2√(3/2) = √(4*3/2) = √6Next, compute sqrt(1 +4*(1/(2√2))²):4*(1/(2√2))² =4*(1/(8)) =0.5Therefore, sqrt(1 +0.5)=sqrt(1.5)=sqrt(3/2)=√6/2Then denominator:16*(1/(2√2))² +1 =16*(1/8) +1=2 +1=3Therefore, A(k)= √6 * (√6/2)/3= (6/2)/3=3/3=1Therefore, maximum area is 1.But wait, original area formula is (1/2)|x1y2 -x2y1|, but according to our calculation, after all substitutions, the Area was expressed as 4√3 k sqrt(1 +4k²)/(16k² +1), and when k=1/(2√2), this expression equals 1.But let me double-check:Wait, A(k) = 4√3 * k * sqrt(1 +4k²)/(16k² +1)k=1/(2√2)Compute each term:4√3 * (1/(2√2)) = 4√3/(2√2) = 2√3/√2 = 2*(√3/√2) = 2*sqrt(3/2) = sqrt(4*(3/2)) = sqrt(6)sqrt(1 +4k²) = sqrt(1 +4*(1/(8)) )= sqrt(1 +0.5)=sqrt(1.5)=sqrt(3/2)=√6/2Denominator:16k² +1=16*(1/8)+1=2 +1=3Multiply together: sqrt(6)*(√6/2)/3 = (6/2)/3=3/3=1Yes, so Area=1.Therefore, the maximum area is 1.But let me check with another approach to verify.Alternatively, we can parametrize H. Since H is the midpoint, and OH=1. Let’s parametrize H as (h_x, h_y) with h_x² + h_y² =1. Then, using the midpoint of a chord of the ellipse. The equation of the chord can be found using the point form.For an ellipse, the equation of the chord with midpoint (h,k) is given by (h x)/a² + (k y)/b² = (h²)/a² + (k²)/b². In our case, a²=4, b²=1.So equation of chord PQ with midpoint H=(h_x,h_y) is (h_x x)/4 + (h_y y)/1 = (h_x²)/4 + (h_y²)/1But since PQ is a chord passing through H, but H is the midpoint.Given that H lies on OH=1, so h_x² + h_y² =1.We can parametrize H as (cosθ, sinθ), since h_x² + h_y²=1.So let H=(cosθ, sinθ). Then the equation of chord PQ is:( cosθ x ) /4 + ( sinθ y ) = (cos²θ)/4 + sin²θSimplify RHS: (cos²θ)/4 + sin²θ = (cos²θ +4 sin²θ)/4Therefore, equation of PQ: (cosθ x)/4 + sinθ y = (cos²θ +4 sin²θ)/4Multiply both sides by 4:cosθ x +4 sinθ y = cos²θ +4 sin²θWe can write this as cosθ x +4 sinθ y = cos²θ +4 sin²θNow, find points P and Q on the ellipse x²/4 + y²=1 satisfying this equation.To find the points of intersection, solve the system:x²/4 + y² =1cosθ x +4 sinθ y = cos²θ +4 sin²θWe can parametrize this.Let me solve for y from the line equation:4 sinθ y = cos²θ +4 sin²θ - cosθ xThus,y = [ cos²θ +4 sin²θ - cosθ x ] / (4 sinθ )Substitute into the ellipse equation:x²/4 + [ ( cos²θ +4 sin²θ - cosθ x ) / (4 sinθ ) ]² =1This seems complicated, but maybe we can use parametric expressions.Alternatively, since P and Q lie on the ellipse, we can write them in parametric form. For an ellipse, parametric equations are x=2cosφ, y=sinφ.So let P=(2cosφ, sinφ), Q=(2cosψ, sinψ). Then, midpoint H is ( (2cosφ +2cosψ)/2, (sinφ + sinψ)/2 )=(cosφ + cosψ, (sinφ + sinψ)/2 )But H is given as (cosθ, sinθ). Therefore,cosφ + cosψ = cosθ(sinφ + sinψ)/2 = sinθSo we have two equations:cosφ + cosψ = cosθsinφ + sinψ = 2 sinθLet me square and add these equations:(cosφ + cosψ)^2 + (sinφ + sinψ)^2 = cos²θ +4 sin²θLeft side:cos²φ + 2cosφcosψ + cos²ψ + sin²φ + 2 sinφ sinψ + sin²ψ = (cos²φ + sin²φ) + (cos²ψ + sin²ψ) +2(cosφcosψ + sinφ sinψ ) =1 +1 +2cos(φ -ψ)=2 +2cos(φ -ψ)Therefore,2 +2cos(φ -ψ)=cos²θ +4 sin²θThus,cos(φ -ψ)= (cos²θ +4 sin²θ -2)/2But let's compute RHS:cos²θ +4 sin²θ -2 = (1 - sin²θ) +4 sin²θ -2 =1 +3 sin²θ -2= -1 +3 sin²θThus,cos(φ -ψ)= (-1 +3 sin²θ)/2But we need to relate this to the area.The area of triangle POQ is (1/2)|x1y2 -x2y1|. For points P=(2cosφ, sinφ) and Q=(2cosψ, sinψ), this becomes:(1/2)|2cosφ sinψ -2cosψ sinφ|= |cosφ sinψ -cosψ sinφ|= |sin(ψ -φ)|Therefore, Area= |sin(ψ -φ)|We need to maximize |sin(ψ -φ)|, which is equivalent to maximizing |sin(Δ)| where Δ=ψ -φ. The maximum value is 1, but we need to see if it's achievable under the given constraints.From the previous equation:cosΔ= (-1 +3 sin²θ)/2So cosΔ= (3 sin²θ -1)/2The maximum value of |sinΔ| is 1, but let's see if this is possible.Given that cosΔ=(3 sin²θ -1)/2, then sinΔ=±sqrt(1 - [ (3 sin²θ -1)/2 ]² )But to have |sinΔ|=1, need [ (3 sin²θ -1)/2 ]²=0So 3 sin²θ -1=0 => sin²θ=1/3 => sinθ=±1/√3Therefore, when sinθ=±1/√3, then cosΔ=0, so Δ=±π/2, so sinΔ=±1. Therefore, Area=1.Therefore, the maximum area is 1.Therefore, this approach also gives maximum area 1. Hence, the answer is 1.But let me check with θ such that sinθ=1/√3.Let’s take θ such that sinθ=1/√3, then cosθ=sqrt(1 -1/3)=sqrt(2/3)=√6/3.Then, midpoint H=(cosθ, sinθ)=(√6/3,1/√3)Then, equation of chord PQ is:cosθ x /4 + sinθ y = (cos²θ)/4 + sin²θPlugging in:(√6/3)x/4 + (1/√3)y = ( (6/9)/4 ) + (1/3 )Simplify:(√6 x)/(12) + y/√3 = ( (2/3)/4 ) +1/3 = (1/6) +1/3=1/2Multiply both sides by 12√3 to eliminate denominators:√6 x * √3 +12 y = 6√3Simplify √6*√3=√18=3√2:3√2 x +12 y =6√3Divide both sides by 3:√2 x +4 y=2√3Now, solve with ellipse equation x²/4 + y²=1Let’s solve for x from the line equation:x= (2√3 -4y)/√2Substitute into ellipse:[ (2√3 -4y)^2 / (4*2) ] + y² =1Compute (2√3 -4y)^2=4*3 -16√3 y +16y²=12 -16√3 y +16y²Divide by 8:(12 -16√3 y +16y²)/8 + y² =1Simplify:(12/8) - (16√3 y)/8 + (16y²)/8 + y²=1= 1.5 -2√3 y +2y² +y²=1.5 -2√3 y +3y²=1Thus,3y² -2√3 y +0.5=0Multiply by 2:6y² -4√3 y +1=0Use quadratic formula:y=(4√3 ±sqrt( (4√3)^2 -4*6*1 ))/(2*6)= (4√3 ±sqrt(48 -24))/12=(4√3 ±sqrt(24))/12=(4√3 ±2√6)/12=(2√3 ±√6)/6Thus, two y-values: y=(2√3 +√6)/6 and y=(2√3 -√6)/6Compute corresponding x:From x=(2√3 -4y)/√2For y=(2√3 +√6)/6:x=(2√3 -4*(2√3 +√6)/6)/√2= (2√3 - (8√3 +4√6)/6)/√2= ( (12√3 -8√3 -4√6)/6 )/√2=( (4√3 -4√6)/6 )/√2= (4(√3 -√6)/6)/√2= (2(√3 -√6)/3)/√2= 2/(3√2)(√3 -√6)= (√3 -√6)/(3*(√2)/2)= wait, perhaps better to rationalize:x=(4√3 -4√6)/(6√2)= (4(√3 -√6))/(6√2)= (2(√3 -√6))/(3√2)Multiply numerator and denominator by √2:2(√3 -√6)√2 / (3*2)= (2√2(√3 -√6))/6= (√2(√3 -√6))/3= (√6 -√12)/3= (√6 -2√3)/3Similarly for the other y:y=(2√3 -√6)/6:x=(2√3 -4*(2√3 -√6)/6)/√2= (2√3 - (8√3 -4√6)/6)/√2= ( (12√3 -8√3 +4√6)/6 )/√2=( (4√3 +4√6)/6 )/√2= 4(√3 +√6)/(6√2)= 2(√3 +√6)/(3√2)= similar to above, multiply by √2/√2:2(√3 +√6)√2/(3*2)= (√2(√3 +√6))/3= (√6 +√12)/3= (√6 +2√3)/3Therefore, points P and Q are:P=( (√6 -2√3)/3 , (2√3 +√6)/6 )Q=( (√6 +2√3)/3 , (2√3 -√6)/6 )Compute the area using determinant formula:Area=(1/2)|x1y2 -x2y1|Compute x1y2 -x2y1:[(√6 -2√3)/3 ] * [ (2√3 -√6)/6 ] - [ (√6 +2√3)/3 ] * [ (2√3 +√6)/6 ]Factor out 1/(3*6)=1/18:[ (√6 -2√3)(2√3 -√6) - (√6 +2√3)(2√3 +√6) ]Compute each product:First term: (√6 -2√3)(2√3 -√6)=√6*2√3 -√6*√6 -2√3*2√3 +2√3*√6=2√18 -6 -4*3 +2√18=2*3√2 -6 -12 +2*3√2=6√2 -18 +6√2=12√2 -18Second term: (√6 +2√3)(2√3 +√6)=√6*2√3 +√6*√6 +2√3*2√3 +2√3*√6=2√18 +6 +12 +2√18=2*3√2 +18 +2*3√2=6√2 +18 +6√2=12√2 +18Therefore,[ (12√2 -18 ) - (12√2 +18) ] /18 = [12√2 -18 -12√2 -18]/18= (-36)/18= -2Therefore, |x1y2 -x2y1|=2, so area=(1/2)*2=1. Which matches our previous result.Therefore, maximum area is indeed 1.Final Answer(1) The standard equation of ellipse ( C ) is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The maximum area of ( triangle POQ ) is boxed{1}.</think>Given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) with an eccentricity of ( frac{sqrt{3}}{2} ) and a point ( left( -sqrt{3}, frac{1}{2} right) ) on the ellipse:(1) To find the standard equation of the ellipse:- The eccentricity ( e = frac{sqrt{3}}{2} ) gives ( c = frac{sqrt{3}}{2}a ).- Using ( c^2 = a^2 - b^2 ), we get ( left( frac{sqrt{3}}{2}a right)^2 = a^2 - b^2 ), leading to ( b^2 = frac{a^2}{4} ).- Substituting the point ( left( -sqrt{3}, frac{1}{2} right) ) into the ellipse equation, we solve for ( a ) and ( b ): [ frac{3}{a^2} + frac{frac{1}{4}}{frac{a^2}{4}} = 1 implies frac{4}{a^2} = 1 implies a^2 = 4 implies a = 2 ] [ b^2 = frac{a^2}{4} = 1 implies b = 1 ]- Thus, the standard equation is ( frac{x^2}{4} + y^2 = 1 ).(2) To find the maximum area of ( triangle POQ ):- The midpoint ( H ) of segment ( PQ ) is at a distance ( OH = 1 ).- Using parametric coordinates and the equation of the chord with midpoint ( H ), we find the maximum area by maximizing the determinant formula for the area.- After solving the system of equations and using calculus, we find that the maximum area occurs when ( k = frac{1}{2sqrt{2}} ).- The maximum area is calculated to be 1 using both parametric and determinant approaches.Final answers:(1) The standard equation of ellipse ( C ) is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The maximum area of ( triangle POQ ) is (boxed{1}).,<think>Okay, let's try to solve this problem step by step. First, part (1) asks for the standard equation of the ellipse C. The given information is that the eccentricity is √3 / 2, and the point (-√3, 1/2) lies on the ellipse. I remember that the standard equation of an ellipse is x²/a² + y²/b² = 1, where a > b > 0. The eccentricity e is given by e = c/a, where c is the distance from the center to each focus. Also, the relationship between a, b, and c is c² = a² - b². Given that e = √3 / 2, so c = (√3 / 2) * a. Then, substituting into c² = a² - b², we get:( (√3 / 2) * a )² = a² - b²Calculating the left side: (3/4)a² = a² - b²So, rearranging gives: b² = a² - (3/4)a² = (1/4)a² => b = a/2So, the semi-minor axis is half of the semi-major axis. Now, since the point (-√3, 1/2) lies on the ellipse, substituting x = -√3 and y = 1/2 into the ellipse equation:( (-√3 )² ) / a² + ( (1/2)² ) / b² = 1Calculating the terms: (3)/a² + (1/4)/b² = 1But we know that b² = (1/4)a², so substituting that in:3/a² + (1/4) / ( (1/4)a² ) = 3/a² + 1/a² = 4/a² = 1Therefore, 4/a² = 1 => a² = 4 => a = 2. Then, since b = a/2, b = 1. So, the standard equation is x²/4 + y²/1 = 1, or x²/4 + y² = 1.Let me check if the point (-√3, 1/2) is on this ellipse:x²/4 = (3)/4, and y² = (1/2)² = 1/4. Adding them: 3/4 + 1/4 = 1. Yep, that works. So part (1) is solved.Now part (2). The problem says that a line l intersects the ellipse C at points P and Q, the midpoint H of PQ is such that OH = 1, where O is the origin. We need to find the maximum area of triangle POQ.First, I need to recall that the area of triangle POQ can be calculated using the determinant formula: 1/2 |x1y2 - x2y1| where P is (x1, y1) and Q is (x2, y2). Alternatively, if we have vectors OP and OQ, the area is 1/2 the magnitude of their cross product. But maybe there's another way. Also, since H is the midpoint of PQ, then H has coordinates ((x1 + x2)/2, (y1 + y2)/2). And OH = 1, so the distance from the origin to H is 1. So, sqrt( [(x1 + x2)/2 ]² + [ (y1 + y2)/2 ]² ) = 1. Squaring both sides: [ (x1 + x2)/2 ]² + [ (y1 + y2)/2 ]² = 1. Also, since P and Q lie on the ellipse x²/4 + y² = 1, so (x1)²/4 + (y1)^2 = 1 and similarly for (x2, y2).We need to maximize the area of triangle POQ, which is 1/2 |x1y2 - x2y1|. Hmm, how to approach this. Maybe parametrize the points P and Q. But since they are intersections of a line with the ellipse, perhaps parametrize the line in terms of its midpoint H. Alternatively, use the concept that for a chord of an ellipse with midpoint H, there's a certain relationship. Also, the condition that OH = 1.Let me recall that for an ellipse, the equation of the chord with midpoint (h, k) is (h x)/a² + (k y)/b² = (h²)/a² + (k²)/b². This is called the equation of the chord with a given midpoint. So, if H is (h, k), then the line PQ is (h x)/4 + (k y)/1 = h²/4 + k². But since H is the midpoint, and OH = 1, we have h² + k² = 1. Wait, no. Wait, OH is the distance from the origin to H, so sqrt(h² + k²) = 1, so h² + k² = 1.So, given that H is on the circle of radius 1 centered at the origin, and H is the midpoint of PQ. The line PQ is given by (h x)/4 + k y = h²/4 + k². Now, points P and Q lie on both the ellipse and this line. To find the area of triangle POQ, we can find the coordinates of P and Q by solving the line equation and the ellipse equation.Alternatively, maybe there's a parametrization approach. Let me think.Alternatively, use vectors. Let me denote vector OP = p and OQ = q. The area is 1/2 |p × q|. Since H is the midpoint, (p + q)/2 = h, where h is a vector with |h| = 1. So p + q = 2h. So, given that p and q are points on the ellipse, and their sum is 2h, where |h| = 1. Then, we need to maximize 1/2 |p × q|.Let me write p = (x1, y1), q = (x2, y2). Then p + q = (2h1, 2h2), where h1² + h2² = 1. Also, x1²/4 + y1² = 1 and x2²/4 + y2² = 1.We need to maximize |x1y2 - x2y1|.Alternatively, perhaps use Lagrange multipliers. But that might be complicated. Maybe there's a more straightforward way.Alternatively, express q = 2h - p. Since p + q = 2h, then q = 2h - p. Since q is on the ellipse, substituting into the ellipse equation:( (2h1 - x1)^2 ) / 4 + (2h2 - y1)^2 = 1But p is also on the ellipse, so x1² /4 + y1² = 1. Let's expand the equation for q:(4h1² - 4h1x1 + x1²)/4 + 4h2² - 4h2y1 + y1² = 1Simplify:( h1² - h1x1 + x1²/4 ) + (4h2² - 4h2y1 + y1² ) = 1But x1²/4 + y1² = 1, so substituting:h1² - h1x1 + 1 + 4h2² - 4h2y1 = 1Simplify:h1² - h1x1 + 4h2² - 4h2y1 = 0But since h1² + h2² = 1 (since |h| = 1), then 4h2² = 4(1 - h1²). So substitute that:h1² - h1x1 + 4(1 - h1²) - 4h2y1 = 0Calculate:h1² - h1x1 + 4 - 4h1² - 4h2y1 = 0Combine like terms:-3h1² - h1x1 - 4h2y1 + 4 = 0Hmm, not sure if this helps. Maybe another approach.Let me recall that in an ellipse, if two points are such that their midpoint is H, then the line PQ has a slope related to the coordinates of H. Specifically, the slope of PQ is - ( (h1)/a² ) / ( (h2)/b² ). In our case, a² = 4, b² = 1, so the slope would be - (h1 / 4 ) / (h2 / 1 ) = - h1 / (4 h2 ). Alternatively, from the chord equation with midpoint H: the slope is - ( (h1)/a² ) / ( (h2)/b² ). So yes, slope m = - (h1 / 4 ) / (h2 ) = - h1 / (4 h2 ). But perhaps we can use parametric equations for the ellipse. Let me parametrize the ellipse. The standard parametrization is x = 2 cos θ, y = sin θ, since a = 2, b = 1. Then, points P and Q can be expressed as (2 cos θ, sin θ) and (2 cos φ, sin φ). Their midpoint H is ( (2 cos θ + 2 cos φ)/2, (sin θ + sin φ)/2 ) = ( cos θ + cos φ, (sin θ + sin φ)/2 ). The distance OH is 1, so sqrt( (cos θ + cos φ)^2 + ( (sin θ + sin φ)/2 )^2 ) = 1. Squaring both sides:(cos θ + cos φ)^2 + ( (sin θ + sin φ)/2 )^2 = 1Let me expand this:cos²θ + 2 cosθ cosφ + cos²φ + ( sin²θ + 2 sinθ sinφ + sin²φ ) / 4 = 1Multiply through by 4 to eliminate the denominator:4 cos²θ + 8 cosθ cosφ + 4 cos²φ + sin²θ + 2 sinθ sinφ + sin²φ = 4Hmm, this seems messy. Maybe use trigonometric identities. Note that cos²θ + sin²θ = 1, same for φ. So:4 (cos²θ + cos²φ ) + 8 cosθ cosφ + (sin²θ + sin²φ ) + 2 sinθ sinφ = 4But cos²θ + cos²φ = (1 - sin²θ) + (1 - sin²φ ) = 2 - (sin²θ + sin²φ )Therefore, substituting:4 [ 2 - (sin²θ + sin²φ ) ] + 8 cosθ cosφ + (sin²θ + sin²φ ) + 2 sinθ sinφ = 4Expand:8 - 4(sin²θ + sin²φ ) + 8 cosθ cosφ + sin²θ + sin²φ + 2 sinθ sinφ = 4Combine like terms:8 - 3(sin²θ + sin²φ ) + 8 cosθ cosφ + 2 sinθ sinφ = 4Bring 8 to the right:-3(sin²θ + sin²φ ) + 8 cosθ cosφ + 2 sinθ sinφ = -4Hmm, this is getting complicated. Maybe there's a better way to parametrize.Alternatively, let's consider that if H is the midpoint, then PQ is a chord with midpoint H. So, the chord can be parametrized in terms of H and the direction of the line. Let me think.Alternatively, since H is on the unit circle, we can parametrize H as (cos α, sin α), since h1² + h2² = 1. Then, the line PQ passes through H and has a certain slope related to the ellipse.From the chord midpoint formula, the slope m of PQ is - (h1 / a² ) / (h2 / b² ) = - (h1 /4 ) / (h2 /1 ) = - h1/(4 h2 ). So, given H = (cos α, sin α), the slope m = - (cos α ) / (4 sin α ). Thus, the line PQ has equation y - sin α = m (x - cos α ). Then, we can find the points P and Q by solving this line equation with the ellipse equation.Once we have P and Q, we can compute the area of triangle POQ.Alternatively, perhaps use parametric coordinates. Let me write the line PQ as passing through H with direction vector. Let me denote t as a parameter. So, points on PQ can be written as H + t*d, where d is the direction vector. But maybe that's more involved.Alternatively, since we have the line equation, substitute into the ellipse and solve for x and y. Let's proceed.Given H = (cos α, sin α ), then the line PQ is:y - sin α = (- cos α / (4 sin α )) (x - cos α )Multiply both sides by 4 sin α to eliminate denominators:4 sin α (y - sin α ) = - cos α (x - cos α )Expand:4 sin α y - 4 sin² α = - cos α x + cos² αBring all terms to one side:cos α x + 4 sin α y = 4 sin² α + cos² αNote that 4 sin² α + cos² α = 3 sin² α + 1.But maybe that's not helpful.So the line equation is cos α x + 4 sin α y = 3 sin² α + 1. Hmm, perhaps substitute into the ellipse equation.The ellipse equation is x² /4 + y² = 1. Let's solve for x and y on the line.From the line equation: cos α x + 4 sin α y = 3 sin² α + 1. Let's solve for y:4 sin α y = 3 sin² α + 1 - cos α x=> y = [ 3 sin² α + 1 - cos α x ] / (4 sin α )Substitute into ellipse equation:x² /4 + [ (3 sin² α + 1 - cos α x ) / (4 sin α ) ]² = 1This seems complicated, but perhaps expand it.Let me denote A = 3 sin² α + 1 - cos α x. Then:x² /4 + (A)^2 / (16 sin² α ) = 1Multiply through by 16 sin² α to eliminate denominators:4 sin² α x² + A² = 16 sin² αBut A = 3 sin² α + 1 - cos α x, so A² is:[3 sin² α + 1]^2 - 2 [3 sin² α + 1] cos α x + cos² α x²Therefore, substituting back:4 sin² α x² + [9 sin^4 α + 6 sin² α + 1 - 6 sin² α cos α x - 2 cos α x + cos² α x²] = 16 sin² αNow, combine like terms:4 sin² α x² + cos² α x² - 6 sin² α cos α x - 2 cos α x + 9 sin^4 α + 6 sin² α + 1 = 16 sin² αGroup x² terms:(4 sin² α + cos² α ) x² - (6 sin² α cos α + 2 cos α ) x + (9 sin^4 α + 6 sin² α + 1 - 16 sin² α ) = 0Simplify each coefficient:First, coefficient of x²:4 sin² α + cos² α = 3 sin² α + (sin² α + cos² α ) = 3 sin² α + 1Coefficient of x:- cos α (6 sin² α + 2 )Constant term:9 sin^4 α + 6 sin² α + 1 - 16 sin² α = 9 sin^4 α - 10 sin² α + 1So, the quadratic equation in x is:(3 sin² α + 1 ) x² - 2 cos α (3 sin² α + 1 ) x + (9 sin^4 α - 10 sin² α + 1 ) = 0Wait, let me check that. The coefficient of x is -cos α (6 sin² α + 2 ). Let me factor:-2 cos α (3 sin² α + 1 )Similarly, the constant term is 9 sin^4 α -10 sin² α +1.So the quadratic equation becomes:(3 sin² α + 1 ) x² - 2 cos α (3 sin² α + 1 ) x + (9 sin^4 α -10 sin² α +1 ) = 0Notice that 3 sin² α + 1 is a common factor in the first two terms. Let me factor that out:(3 sin² α + 1 ) [ x² - 2 cos α x ] + (9 sin^4 α -10 sin² α +1 ) = 0But not sure if that's helpful. Alternatively, let me denote t = sin² α. Then, cos² α = 1 - t.But maybe not.Alternatively, since this is a quadratic in x, the solutions for x will give us the x-coordinates of points P and Q. The sum of roots (x1 + x2) is [ 2 cos α (3 sin² α +1 ) ] / (3 sin² α +1 ) ) = 2 cos α. So x1 + x2 = 2 cos α, which makes sense because the midpoint H has x-coordinate cos α.Similarly, the product of roots x1 x2 = [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 )But since H is the midpoint, we know x1 + x2 = 2 cos α, so the average is cos α, which is correct.Now, since we need to find the area of triangle POQ, which is 1/2 |x1 y2 - x2 y1 |. Since y1 and y2 can be expressed in terms of x1 and x2 using the line equation. Alternatively, maybe express y1 and y2 in terms of α. Let's recall that for points P and Q on the line PQ, which has equation y = [3 sin² α +1 - cos α x ] / (4 sin α )Thus, y1 = [3 sin² α +1 - cos α x1 ] / (4 sin α )Similarly, y2 = [3 sin² α +1 - cos α x2 ] / (4 sin α )Therefore, the area is 1/2 |x1 y2 - x2 y1 | = 1/2 |x1 [ (3 sin² α +1 - cos α x2 ) / (4 sin α ) ] - x2 [ (3 sin² α +1 - cos α x1 ) / (4 sin α ) ] | Simplify:= 1/(8 sin α ) |x1 (3 sin² α +1 ) - x1 x2 cos α - x2 (3 sin² α +1 ) + x1 x2 cos α |Notice that the terms with x1 x2 cos α cancel:= 1/(8 sin α ) | (3 sin² α +1 )(x1 - x2 ) |= (3 sin² α +1 )/(8 sin α ) |x1 - x2|Therefore, the area is (3 sin² α +1 )/(8 sin α ) |x1 - x2|Now, |x1 - x2| can be found from the quadratic equation. For a quadratic ax² + bx + c =0, the difference of roots is sqrt( (x1 +x2 )² -4x1x2 ). But actually, |x1 - x2| = sqrt( (x1 +x2 )² -4x1x2 )We have x1 + x2 = 2 cos α, and x1 x2 = [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 )Thus,|x1 - x2| = sqrt( (2 cos α )² - 4 * [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 ) )Simplify:= sqrt(4 cos² α - 4 [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 ) )Factor out 4:= 2 sqrt( cos² α - [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 ) )Let me compute the expression inside the sqrt:cos² α - [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 )First, write cos² α as 1 - sin² α:= (1 - sin² α ) - [9 sin^4 α -10 sin² α +1 ] / (3 sin² α +1 )Let me combine these terms:= [ (1 - sin² α )(3 sin² α +1 ) - 9 sin^4 α +10 sin² α -1 ] / (3 sin² α +1 )Expand the numerator:(1)(3 sin² α ) + (1)(1) - sin² α (3 sin² α ) - sin² α (1 ) -9 sin^4 α +10 sin² α -1Calculate term by term:3 sin² α +1 -3 sin^4 α - sin² α -9 sin^4 α +10 sin² α -1Combine like terms:3 sin² α - sin² α +10 sin² α = 12 sin² α1 -1 = 0-3 sin^4 α -9 sin^4 α = -12 sin^4 αSo numerator becomes 12 sin² α -12 sin^4 α = 12 sin² α (1 - sin² α ) = 12 sin² α cos² αTherefore, the expression inside the sqrt is:[12 sin² α cos² α ] / (3 sin² α +1 )Thus, |x1 -x2| = 2 sqrt( [12 sin² α cos² α ] / (3 sin² α +1 ) )= 2 * [ sqrt(12 ) sin α cos α ] / sqrt(3 sin² α +1 )Simplify sqrt(12 ) = 2 sqrt(3 ), so:= 2 * [ 2 sqrt(3 ) sin α cos α ] / sqrt(3 sin² α +1 )= 4 sqrt(3 ) sin α cos α / sqrt(3 sin² α +1 )Therefore, going back to the area:Area = (3 sin² α +1 )/(8 sin α ) * [4 sqrt(3 ) sin α cos α / sqrt(3 sin² α +1 ) ]Simplify:The sin α in the denominator cancels with the sin α in the numerator:= (3 sin² α +1 )/(8 ) * [4 sqrt(3 ) cos α / sqrt(3 sin² α +1 ) ]Multiply constants:4/8 = 1/2, so:= (3 sin² α +1 )/(2 ) * [ sqrt(3 ) cos α / sqrt(3 sin² α +1 ) ]Note that (3 sin² α +1 ) / sqrt(3 sin² α +1 ) = sqrt(3 sin² α +1 )Thus, Area = ( sqrt(3 sin² α +1 ) / 2 ) * sqrt(3 ) cos α= ( sqrt(3 ) / 2 ) * sqrt(3 sin² α +1 ) * cos αWait, let me check:Wait, the expression is [ (3 sin² α +1 ) / sqrt(3 sin² α +1 ) ] * [ sqrt(3 ) cos α / 2 ]Which simplifies to sqrt(3 sin² α +1 ) * sqrt(3 ) cos α / 2= sqrt(3 ) cos α sqrt(3 sin² α +1 ) /2But sqrt(3 ) * sqrt(3 sin² α +1 ) = sqrt( 3 (3 sin² α +1 ) )Wait, no. sqrt(a) * sqrt(b) = sqrt(ab ). So:sqrt(3 ) * sqrt(3 sin² α +1 ) = sqrt( 3 (3 sin² α +1 ) )Thus, Area = [ sqrt( 3 (3 sin² α +1 ) ) * cos α ] / 2= [ sqrt( 9 sin² α +3 ) * cos α ] /2= [ sqrt(9 sin² α +3 ) * cos α ] /2Hmm, maybe we can factor out 3 inside the sqrt:sqrt( 3 (3 sin² α +1 ) )Wait, but we have:sqrt( 3 (3 sin² α +1 ) ) * cos α / 2 = sqrt(3) * sqrt(3 sin² α +1 ) * cos α /2Hmm, not sure. Let's keep it as sqrt(9 sin² α +3 ) * cos α /2. Alternatively, write sqrt(9 sin² α +3 ) = sqrt(3(3 sin² α +1 )). But maybe square the area to maximize it.Let me denote A = Area. Then,A = [ sqrt(9 sin² α +3 ) * cos α ] /2Thus, A² = (9 sin² α +3 ) * cos² α /4We need to maximize A², which is equivalent to maximizing A.So, maximize (9 sin² α +3 ) * cos² αLet me set t = sin² α, so cos² α =1 - t. Then, the expression becomes (9 t +3 )(1 - t ) = 3(3 t +1 )(1 - t )Expand:3 [ 3 t (1 - t ) +1 (1 - t ) ] = 3 [ 3 t -3 t² +1 -t ] = 3 [ 2 t -3 t² +1 ]But maybe just compute the original expansion:(9 t +3 )(1 - t ) = 9 t (1 - t ) +3 (1 - t ) =9 t -9 t² +3 -3 t =6 t -9 t² +3Thus, A² = [6 t -9 t² +3 ] / (4 )So we need to maximize f(t ) = -9 t² +6 t +3, where t = sin² α ∈ [0,1].This is a quadratic function in t, opening downward (since coefficient of t² is negative). The maximum occurs at vertex. The vertex is at t = -b/(2a ) = -6/(2*(-9 )) = 6/(18 )= 1/3.Thus, maximum f(t ) at t=1/3. Compute f(1/3 ):-9*(1/3 )² +6*(1/3 ) +3 = -9*(1/9 ) + 2 +3 = -1 +5=4Therefore, maximum A²=4/4=1 => maximum A=1.Wait, that can't be right. Wait, f(t ) maximum is 4, then A²=4/4=1, so A=1. But is this correct? Let me check.Wait, if f(t )= -9 t² +6 t +3, then at t=1/3, f(t )= -9*(1/9 ) +6*(1/3 ) +3= -1 +2 +3=4. Therefore, A²=4/4=1, so A=1. So maximum area is 1.But is this correct? Let me verify with a specific case. For example, if α=0, then H=(1,0). Then, line PQ is the horizontal line through (1,0). The ellipse at x=1, y=0 is the point (2*1,0)=(2,0)? Wait, no. Wait, H is (cos 0, sin 0 )=(1,0). Then, the line PQ is horizontal line y=0. The ellipse x²/4 + y²=1 intersects y=0 at x=±2. So points P=(2,0), Q=(-2,0). Midpoint H=(0,0 ), but wait, midpoint of (2,0 ) and (-2,0 ) is (0,0 ), but H was supposed to be (1,0 ). Contradiction. Therefore, my parametrization must be wrong. Wait, this suggests a problem with my approach.Wait, when I parametrized H as (cos α, sin α ), but in reality, when H is (1,0 ), which is on the unit circle, but the midpoint of PQ should be (1,0 ). However, the line PQ would be the horizontal line y=0, which intersects the ellipse at x=±2, y=0. The midpoint of those points is (0,0 ), not (1,0 ). So this is a contradiction. Hence, my parametrization is flawed.Wait, what's wrong here. I think the error comes from the chord with midpoint H. If H is outside the ellipse, then the chord might not exist. But in our case, H is on the unit circle, which is inside the ellipse since the ellipse has semi-major axis 2. So, points on the unit circle are inside the ellipse. But in the example above, H=(1,0 ), which is inside the ellipse, but the line through H with the appropriate slope would intersect the ellipse at two points P and Q whose midpoint is H. However, when I tried with H=(1,0 ), the line PQ would actually need to have points whose midpoint is (1,0 ). But when I used the chord equation, I obtained the line equation leading to midpoint (0,0 ), which is a contradiction. So my mistake must be in assuming H=(cos α, sin α ). Wait, no. The chord equation with midpoint H is correct, but if the line PQ has midpoint H=(h1, h2 ), then solving the line and ellipse equation should give points P and Q with midpoint H. But in the case of H=(1,0 ), the line PQ obtained from the chord equation is x/4 +0= (1)/4 +0 => x=1. So the vertical line x=1. But x=1 intersects the ellipse at y²=1 -1/4=3/4, so y=±√3/2. Thus, points P=(1, √3/2 ), Q=(1, -√3/2 ), midpoint H=(1,0 ), which is correct. Then, the area of triangle POQ would be 1/2 |x1 y2 -x2 y1 | =1/2 |1*(-√3/2 ) -1*(√3/2 )| =1/2 | -√3/2 -√3/2 | =1/2 | -√3 | = √3 /2 ≈0.866. But according to our previous conclusion, maximum area is 1. So this example gives area less than 1, so maybe the maximum area is indeed 1.Wait, let's check another case. Suppose H is at (0,1 ). Then the line PQ has midpoint (0,1 ). Using the chord equation, the slope is - (0 /4 ) / (1 /1 )=0. So horizontal line. The line is y=1. Substituting into the ellipse equation: x²/4 +1=1 =>x²/4=0 =>x=0. So the line intersects the ellipse only at (0,1 ), so it's a tangent. But midpoint is the same point, so this is a degenerate case. Thus, this H=(0,1 ) is not possible because the line would be tangent, so there's no chord. Therefore, H=(0,1 ) is not a valid midpoint. Thus, our parametrization H=(cos α, sin α ) includes points where the chord is valid (i.e., not tangent). So, in order for the line to intersect the ellipse at two distinct points, the discriminant of the quadratic equation must be positive. But in our previous calculation, we arrived at maximum area=1. Let's check with another example. Suppose α=π/6, so H=(cos π/6, sin π/6 )=(√3/2, 1/2 ). The slope of PQ is - (cos α )/(4 sin α ) = - (√3/2 )/(4*(1/2 )) = - (√3/2 )/2 = -√3 /4. So the line PQ is y -1/2 = (-√3 /4 )(x - √3 /2 ). Let's find points P and Q.Substitute into ellipse equation:x²/4 + y² =1.Express y from the line equation:y = (-√3 /4 )x + (√3 /4 )(√3 /2 ) +1/2 = (-√3 /4 )x + (3/8 ) +1/2 = (-√3 /4 )x +7/8.Substitute into ellipse:x²/4 + [ (-√3 /4 x +7/8 ) ]² =1This will give quadratic in x. Let me compute:First, square the y term:= ( ( -√3 x /4 +7/8 ) )² = (3x²)/16 - (2*√3 x*7)/(4*8 ) +49/64 = 3x²/16 - (14√3 x)/32 +49/64Thus, the ellipse equation becomes:x²/4 + 3x²/16 -14√3 x /32 +49/64 =1Multiply all terms by 64 to eliminate denominators:16x² +12x² -28√3 x +49 =64Combine like terms:28x² -28√3 x +49 -64 =0 =>28x² -28√3 x -15=0Divide by GCD (if possible). Not sure. Solve for x using quadratic formula:x = [28√3 ± sqrt( (28√3 )² -4*28*(-15 ) ) ]/(2*28 )Compute discriminant:(28√3 )² +4*28*15= 784*3 + 1680= 2352 +1680=4032sqrt(4032)=sqrt(16*252 )=4*sqrt(252 )=4*sqrt(36*7 )=4*6√7=24√7Thus, x= [28√3 ±24√7 ]/(56 )= [28√3 ±24√7 ]/56 = [7√3 ±6√7 ]/14Therefore, two x-coordinates. Then, y is computed from y = (-√3 /4 )x +7/8.Compute y1 and y2:y1 = (-√3 /4 )( (7√3 +6√7 )/14 ) +7/8= (-√3 (7√3 +6√7 ) )/(56 ) +7/8= ( -7*3 -6√21 )/56 +7/8= ( -21 -6√21 )/56 +49/56= (28 -6√21 )/56 = (14 -3√21 )/28Similarly, y2 = (-√3 /4 )( (7√3 -6√7 )/14 ) +7/8= (-√3 (7√3 -6√7 ) )/(56 ) +7/8= ( -7*3 +6√21 )/56 +49/56= ( -21 +6√21 )/56 +49/56= (28 +6√21 )/56 = (14 +3√21 )/28Therefore, points P and Q are:P = ( (7√3 +6√7 )/14, (14 -3√21 )/28 )Q = ( (7√3 -6√7 )/14, (14 +3√21 )/28 )Now, compute the area of triangle POQ:1/2 |x1 y2 -x2 y1 |Compute x1 y2:[(7√3 +6√7 )/14 ] * [ (14 +3√21 )/28 ]x2 y1:[(7√3 -6√7 )/14 ] * [ (14 -3√21 )/28 ]Subtract them and take absolute value:| [ (7√3 +6√7 )(14 +3√21 ) - (7√3 -6√7 )(14 -3√21 ) ] / (14*28 ) | *1/2Compute numerator:First term: (7√3 +6√7 )(14 +3√21 )=7√3 *14 +7√3 *3√21 +6√7 *14 +6√7 *3√21=98√3 +21√{3*21} +84√7 +18√{7*21}=98√3 +21√{63} +84√7 +18√{147}Simplify radicals:√63=3√7, √147=7√3Thus:98√3 +21*3√7 +84√7 +18*7√3=98√3 +63√7 +84√7 +126√3Combine like terms:(98√3 +126√3 ) + (63√7 +84√7 )=224√3 +147√7Second term: (7√3 -6√7 )(14 -3√21 )=7√3 *14 -7√3 *3√21 -6√7 *14 +6√7 *3√21=98√3 -21√{63} -84√7 +18√{147}Simplify radicals:=98√3 -21*3√7 -84√7 +18*7√3=98√3 -63√7 -84√7 +126√3Combine like terms:(98√3 +126√3 ) + (-63√7 -84√7 )=224√3 -147√7Thus, numerator:[224√3 +147√7 - (224√3 -147√7 ) ] =224√3 +147√7 -224√3 +147√7=294√7Thus, the absolute value is |294√7 / (14*28 ) | *1/2Compute denominator:14*28=392Thus, 294√7 /392 = (294/392 )√7 = (3/4 )√7Multiply by 1/2: (3/4 )√7 *1/2= (3/8 )√7 ≈0. But according to our previous result, the maximum area is 1, but here, the area is (3/8 )√7≈0.99. Close to 1, but not exactly. Maybe due to calculation approximation. Wait, 3/8 *√7 ≈3/8*2.6458≈3*0.3307≈0.992. Which is approximately 1. So maybe the maximum area is indeed 1, achieved when the area expression we derived reaches 1. Therefore, my earlier conclusion seems plausible.Alternatively, maybe there is a case where the area is exactly 1. Let's try when H is at (0,0 ), but OH=1 is not satisfied. H must be OH=1, so H is on the unit circle. Wait, if H is (0,1 ), but that gives a degenerate line. So perhaps when α is such that the expression A=1.Alternatively, maybe set Area=1 and solve for α. From the expression A= [ sqrt(9 sin² α +3 ) * cos α ] /2 =1Thus, sqrt(9 sin² α +3 ) * cos α =2Square both sides: (9 sin² α +3 ) cos² α =4Let t = sin² α. Then, cos² α =1 -t.Thus, (9t +3 )(1 -t ) =4Which is the same as the previous equation: -9t² +6 t +3=4 => -9t² +6 t -1=0 =>9t² -6 t +1=0Solutions: t=(6±sqrt(36 -36 ))/18=6/18=1/3.Thus, t=1/3, so sin² α=1/3, so sin α=±1/√3, cos α=±sqrt(2/3 ). Thus, H=(sqrt(2/3 ), 1/√3 ) or similar. Therefore, at this angle α, the area is indeed 1. Therefore, the maximum area is 1.Thus, after a long process, the maximum area of triangle POQ is 1.Final Answer(1) The standard equation of ellipse ( C ) is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The maximum area of ( triangle POQ ) is boxed{1}.</think>Given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) with an eccentricity of ( frac{sqrt{3}}{2} ) and a point ( left( -sqrt{3}, frac{1}{2} right) ) on the ellipse:1. Finding the standard equation of ellipse ( C ): - The eccentricity ( e = frac{sqrt{3}}{2} ) gives ( c = frac{sqrt{3}}{2}a ). - Using ( c^2 = a^2 - b^2 ), we get ( left( frac{sqrt{3}}{2}a right)^2 = a^2 - b^2 ), leading to ( b^2 = frac{1}{4}a^2 ). - Substituting the point ( left( -sqrt{3}, frac{1}{2} right) ) into the ellipse equation: [ frac{(-sqrt{3})^2}{a^2} + frac{left( frac{1}{2} right)^2}{left( frac{a}{2} right)^2} = 1 implies frac{3}{a^2} + frac{1}{4} cdot frac{4}{a^2} = 1 implies frac{4}{a^2} = 1 implies a^2 = 4 implies a = 2 ] - Therefore, ( b = 1 ), and the standard equation is ( frac{x^2}{4} + y^2 = 1 ).2. Finding the maximum area of ( triangle POQ ): - The midpoint ( H ) of segment ( PQ ) satisfies ( OH = 1 ), implying ( H ) lies on the unit circle. - Using the chord equation with midpoint ( H ), the line ( PQ ) intersects the ellipse at points ( P ) and ( Q ). - The area of ( triangle POQ ) is maximized by considering the parametrization of ( H ) on the unit circle and solving the resulting quadratic equations. - After detailed calculations, the maximum area is found to be 1 by maximizing the expression involving trigonometric identities and quadratic terms.Final answers:1. The standard equation of ellipse ( C ) is (boxed{dfrac{x^2}{4} + y^2 = 1}).2. The maximum area of (triangle POQ) is (boxed{1}).]

🔑:(1) From the given information, we have frac{c}{a} = frac{sqrt{3}}{2} and frac{3}{a^{2}} + frac{frac{1}{4}}{b^{2}} = 1.Solving these equations, we get a^{2}=4 and b^{2}=1.Thus, the equation of ellipse C is frac{x^{2}}{4}+y^{2}=1.(2) Let the intersection of line l with the x-axis be D(n,0), and the equation of line l is x=my+n. The points of intersection of line l with the ellipse are P(x_{1},y_{1}) and Q(x_{2},y_{2}).Solving the system of equations begin{cases} x=my+n, frac{x^{2}}{4}+y^{2}=1, end{cases} we get (4+m^{2})y^{2}+2mny+n^{2}-4=0.Then, y_{1,2}=frac{-2mn pm sqrt{(2mn)^{2}-4(4+m^{2})(n^{2}-4)}}{2(4+m^{2})}.So, frac{y_{1}+y_{2}}{2} = -frac{mn}{4+m^{2}} and y_{1}y_{2}=frac{n^{2}-4}{4+m^{2}}.Thus, frac{x_{1}+x_{2}}{2} = frac{m(y_{1}+y_{2})+2n}{2} = frac{4n}{4+m^{2}},and Hleft( frac{4n}{4+m^{2}},-frac{mn}{4+m^{2}} right).From OH=1, we have n^{2}=frac{(4+m^{2})^{2}}{16+m^{2}}.Then, S_{triangle POQ}=frac{1}{2} OD cdot |y_{1}-y_{2}|=frac{1}{2} |n| cdot |y_{1}-y_{2}|.Let T=n^{2}(y_{1}-y_{2})^{2}=n^{2}[(y_{1}+y_{2})^{2}-4y_{1}y_{2}]=12 cdot 16 cdot frac{4+m^{2}}{(16+m^{2})^{2}}.Set t=4+m^{2} (t geqslant 4).Then, frac{4+m^{2}}{(16+m^{2})^{2}} = frac{t}{t^{2}+24t+144} = frac{1}{t+frac{144}{t}+24} leqslant frac{1}{48}.The equality holds if and only if t=frac{144}{t}, i.e., t=12.Thus, the maximum area of triangle POQ is boxed{1}.

🧠:[<think>Okay, so I need to find sin(α + β) given that cos α cos β = -1. Hmm, let's start by recalling some trigonometric identities that might be useful here. I remember that the cosine of a sum or difference can be expressed as cos α cos β minus or plus sin α sin β. But here we have a product of cosines equal to -1, which is interesting because the maximum value of cosine is 1 and the minimum is -1. So, for their product to be -1, each cosine must be either 1 or -1. Wait, let me check that.If cos α cos β = -1, then since the maximum of each cosine is 1 and the minimum is -1, the only way their product is -1 is if one of them is 1 and the other is -1. Because 1 * (-1) = -1, and (-1) * 1 = -1. So that means either cos α = 1 and cos β = -1, or cos α = -1 and cos β = 1. Let me note that down.Case 1: cos α = 1 and cos β = -1.If cos α = 1, then α must be an angle where cosine is 1, which is α = 2πk, where k is an integer. Similarly, if cos β = -1, then β must be π + 2πm, where m is an integer. So, α = 2πk and β = π + 2πm.Then, sin(α + β) = sin(2πk + π + 2πm) = sin(π + 2π(k + m)). The sine of π plus any multiple of 2π is sin π = 0. Because sin(π + x) = -sin x, but since x here is 2π(k + m), which is a multiple of 2π, sin(π + 2π(k + m)) = -sin(2π(k + m)) = -0 = 0. So in this case, sin(α + β) = 0.Case 2: cos α = -1 and cos β = 1.Similarly, here α = π + 2πk and β = 2πm. Then, α + β = π + 2πk + 2πm = π + 2π(k + m). Again, sin(π + 2π(k + m)) = sin π = 0, same reasoning as before.So in both cases, sin(α + β) = 0.Wait, but is there another possibility? Let me think. What if both cos α and cos β are equal to -1? Then their product would be 1, which doesn't satisfy the equation. Similarly, if both are 1, their product is 1. So the only way the product is -1 is if one is 1 and the other is -1, as I thought earlier.Therefore, regardless of the specific angles α and β, as long as their cosine product is -1, their sum will have a sine of 0. So the answer is 0.But wait, let me verify with another approach. Maybe using identities. Let's recall that sin(α + β) = sin α cos β + cos α sin β.But we know cos α cos β = -1. If I can find sin α and sin β, then maybe plug into this identity. But how?If cos α is 1, then sin α is 0, because sin²α + cos²α = 1. So if cos α = 1, then sin α = 0. Similarly, if cos β = -1, then sin β = 0. Then, sin(α + β) = 0 * (-1) + 1 * 0 = 0 + 0 = 0. Similarly, if cos α = -1 and cos β = 1, then sin α = 0 and sin β = 0. So again, sin(α + β) = (-1)*0 + 0*0 = 0. Wait, no, actually, the formula is sin α cos β + cos α sin β.Wait, let's take case 1: cos α = 1 and cos β = -1. Then sin α = sqrt(1 - cos²α) = sqrt(1 - 1) = 0. Similarly, sin β = sqrt(1 - cos²β) = sqrt(1 - 1) = 0. Therefore, sin(α + β) = sin α cos β + cos α sin β = 0*(-1) + 1*0 = 0 + 0 = 0.Case 2: cos α = -1 and cos β = 1. Similarly, sin α = sqrt(1 - (-1)^2) = 0, and sin β = sqrt(1 - 1^2) = 0. Then, sin(α + β) = 0*1 + (-1)*0 = 0 + 0 = 0.Therefore, either way, sin(α + β) = 0. So that's consistent with my previous conclusion.Alternatively, another approach: if cos α cos β = -1, then perhaps α and β are angles such that α + β is a multiple of π? Let's see. If α is 0 (cos α = 1) and β is π (cos β = -1), then α + β = π, sin π = 0. If α is π and β is 0, same result. If α is 2π and β is π, same thing. So sin(π + 2πk) is always 0.So regardless of the specific angles, as long as their cosines multiply to -1, their sum is an odd multiple of π, whose sine is 0. Hence, sin(α + β) = 0.Therefore, the answer is 0.Final AnswerThe value of sin (alpha + beta) is boxed{0}.</think>Given that (cos alpha cos beta = -1), we need to find (sin (alpha + beta)).First, we note that the product (cos alpha cos beta = -1) implies that one of the cosines must be 1 and the other must be -1. This is because the maximum and minimum values of the cosine function are 1 and -1, respectively, and their product can only be -1 if one is 1 and the other is -1.Case 1: (cos alpha = 1) and (cos beta = -1)- If (cos alpha = 1), then (alpha = 2kpi) for some integer (k), and (sin alpha = 0).- If (cos beta = -1), then (beta = pi + 2mpi) for some integer (m), and (sin beta = 0).- Therefore, (alpha + beta = 2kpi + pi + 2mpi = pi + 2(k + m)pi).- The sine of this angle is (sin(pi + 2(k + m)pi) = sin(pi) = 0).Case 2: (cos alpha = -1) and (cos beta = 1)- If (cos alpha = -1), then (alpha = pi + 2kpi) for some integer (k), and (sin alpha = 0).- If (cos beta = 1), then (beta = 2mpi) for some integer (m), and (sin beta = 0).- Therefore, (alpha + beta = pi + 2kpi + 2mpi = pi + 2(k + m)pi).- The sine of this angle is (sin(pi + 2(k + m)pi) = sin(pi) = 0).In both cases, we find that (sin(alpha + beta) = 0).Using the identity (sin(alpha + beta) = sin alpha cos beta + cos alpha sin beta), we substitute (sin alpha = 0) and (sin beta = 0) in both cases, which confirms that (sin(alpha + beta) = 0).Thus, the value of (sin (alpha + beta)) is (boxed{0}).,<think>Okay, so I need to find sin(α + β) given that cos α cos β = -1. Hmm, let's see. I remember some trigonometric identities that might help here. The problem gives me the product of cosines equals -1, and I need to find the sine of the sum of two angles. First, maybe I can recall the formula for sin(α + β). Let me write that down: sin(α + β) = sin α cos β + cos α sin β. Right, so that's the identity. But I don't know sin α or sin β individually, just the product of the cosines. Hmm. Maybe there's a way to relate this product to something else?Given that cos α cos β = -1. Well, the maximum value of cos α and cos β is 1, and the minimum is -1. So if their product is -1, that means one of them must be 1 and the other -1, right? Because 1 * (-1) = -1. If both were 1 or both were -1, the product would be 1. So that tells us that one cosine is 1 and the other is -1. Wait, but is that the only possibility? Let me check. Suppose cos α = 1, then cos β must be -1. Alternatively, if cos α = -1, then cos β must be 1. Are there any other possibilities where the product is -1? Well, if the cosines are real numbers, then the maximum absolute value of their product is 1, since each cosine is between -1 and 1. So to get exactly -1, they need to be at the extremes. So yes, one is 1 and the other is -1. So that must mean that α is 0 or π, and β is π or 0, but adjusted for periodicity. But maybe we can just consider the principal values here.So if cos α = 1, then α must be 0 (or 2π, 4π, etc.), and cos β = -1, which would mean β is π (or π + 2πk). Similarly, if cos α = -1, then α is π, and cos β = 1, so β is 0. So in either case, α and β are angles separated by π radians. So let's take the first case: α = 0, β = π. Then sin(α + β) = sin(0 + π) = sin(π) = 0. Second case: α = π, β = 0. Then sin(π + 0) = sin(π) = 0. Wait, same result. So in both cases, sin(α + β) is zero. Alternatively, let's think if there's another way to approach this, maybe using identities. Since we know that cos α cos β = -1, perhaps we can use the identity for cos(α + β) or cos(α - β). But the problem is asking for sin(α + β), so maybe combining with other equations.Wait, we know that cos(α + β) = cos α cos β - sin α sin β. And we know cos α cos β = -1. So cos(α + β) = -1 - sin α sin β. But I don't know sin α sin β. Hmm. Not sure if that helps.Alternatively, since we already deduced that either cos α =1 and cos β=-1 or vice versa, then we can compute sin α and sin β. If cos α =1, then sin α = sqrt(1 - cos² α) = sqrt(1 -1)=0. Similarly, if cos β=-1, then sin β = sqrt(1 - (-1)^2) = sqrt(1 -1)=0. So in either case, both sin α and sin β would be zero. Therefore, sin(α + β) = sin α cos β + cos α sin β = 0 * (-1) + 1 * 0 = 0. Similarly in the other case, same result.Therefore, regardless of which angle is which, sin(α + β) will be zero.Alternatively, maybe we can think of it algebraically. If cos α cos β = -1, then since cos α and cos β are both between -1 and 1, the only way their product is -1 is if one is 1 and the other is -1. So without loss of generality, let’s say cos α = 1 and cos β = -1. Then α = 2πk and β = π + 2πm, where k and m are integers. Then α + β = 2πk + π + 2πm = π(2k + 1 + 2m). So sin(α + β) = sin(π(2k + 1 + 2m)) = sin(π + 2π(k + m)) = sin(π) = 0. Similarly, if cos α = -1 and cos β = 1, then α = π + 2πk and β = 2πm. Then α + β = π + 2πk + 2πm = π + 2π(k + m). So sin(α + β) = sin(π + 2π(k + m)) = sin(π) = 0. Therefore, in all cases, sin(α + β) must be 0.Alternatively, let me try another approach. Suppose we square both sides of some equation. Wait, but we only have cos α cos β = -1. If we square it, we get cos²α cos²β = 1. But since cos²θ ≤ 1 for any θ, equality holds only if cos²α = 1 and cos²β = 1. Therefore, cos α and cos β must each be either 1 or -1, which is the same conclusion as before.Therefore, regardless of the approach, the result is sin(α + β) = 0. I think that's solid. So the answer is 0.Final Answerboxed{0}</think>Given that (cos alpha cos beta = -1), we need to find (sin (alpha + beta)).First, we note that the maximum value of (cos alpha) and (cos beta) is 1, and the minimum value is -1. For their product to be -1, one of (cos alpha) or (cos beta) must be 1 and the other must be -1. This implies that either (cos alpha = 1) and (cos beta = -1) or (cos alpha = -1) and (cos beta = 1).If (cos alpha = 1), then (alpha = 2kpi) for some integer (k), and if (cos beta = -1), then (beta = pi + 2mpi) for some integer (m). Similarly, if (cos alpha = -1), then (alpha = pi + 2kpi) for some integer (k), and if (cos beta = 1), then (beta = 2mpi) for some integer (m).In both cases, we calculate (alpha + beta):- If (alpha = 2kpi) and (beta = pi + 2mpi), then (alpha + beta = 2kpi + pi + 2mpi = pi + 2(k + m)pi).- If (alpha = pi + 2kpi) and (beta = 2mpi), then (alpha + beta = pi + 2kpi + 2mpi = pi + 2(k + m)pi).In both scenarios, (sin(alpha + beta) = sin(pi + 2(k + m)pi) = sin(pi) = 0).Thus, the value of (sin (alpha + beta)) is (boxed{0}).]

🔑:From the given information and the fact that |cos alpha| leqslant 1 and |cos beta| leqslant 1, we can deduce that one of cos alpha and cos beta must be 1 and the other must be -1. By considering different cases, we can find the value of alpha + beta, which will then allow us to find sin (alpha + beta).We will consider two cases:1. cos alpha = 1 and cos beta = -12. cos alpha = -1 and cos beta = 1For case 1, we have alpha = 2npi and beta = (2m+1)pi for some integers n and m. Therefore,alpha + beta = 2npi + (2m+1)pi = (2n + 2m + 1)piNow, let's find sin(alpha + beta):sin(alpha + beta) = sin((2n + 2m + 1)pi) = sin(pi) = 0For case 2, we have alpha = (2n+1)pi and beta = 2mpi. Therefore,alpha + beta = (2n+1)pi + 2mpi = (2n + 2m + 1)piAgain, let's find sin(alpha + beta):sin(alpha + beta) = sin((2n + 2m + 1)pi) = sin(pi) = 0Thus, for both cases, we have boxed{sin(alpha + beta) = 0}.

🧠:[<think>Okay, so I need to find the smallest natural number that when multiplied by 7 gives a number made up entirely of nines. Hmm, let's see. Let me start by understanding the problem better.First, the question is asking for the smallest natural number, let's call it 'k', such that 7 multiplied by k equals a number that's all nines. For example, numbers like 9, 99, 999, 9999, etc. So, mathematically, we can write this as:7 * k = 999...9 (some number of nines)Our goal is to find the smallest such k. Alright, so maybe I can approach this by considering numbers consisting of nines and see which one is divisible by 7. The first number that is divisible by 7 will give me the corresponding k when divided by 7.Let me start testing with the smaller numbers made up of nines:- 9 divided by 7: 9 ÷ 7 = 1.285... Not an integer.- 99 divided by 7: 99 ÷ 7 ≈ 14.142... Still not an integer.- 999 ÷ 7: Let's calculate that. 7*142 = 994, so 999 - 994 = 5. Remainder 5. Not divisible.- 9999 ÷ 7: Let me divide step by step. 7*1428 = 9996. Then 9999 - 9996 = 3. Remainder 3. Not divisible.- 99999 ÷ 7: Hmm, 7*14285 = 99995. Subtract that from 99999: 99999 - 99995 = 4. Remainder 4. Not good.- 999999 ÷ 7: Let me compute this. 7*142857 = 999999. Wait, 142857*7? Let me check:142857 * 7:7 * 100000 = 7000007 * 40000 = 280000 → 700000 + 280000 = 9800007 * 2000 = 14000 → 980000 + 14000 = 9940007 * 800 = 5600 → 994000 + 5600 = 9996007 * 50 = 350 → 999600 + 350 = 9999507 * 7 = 49 → 999950 + 49 = 999999Wow, that works! So 7 multiplied by 142857 equals 999999. Therefore, the smallest such number is 142857, right? But let me confirm that there isn't a smaller number of nines that is divisible by 7. Let's recap:- 9: 1 nine, remainder 2- 99: 2 nines, remainder 1- 999: 3 nines, remainder 5- 9999: 4 nines, remainder 3- 99999: 5 nines, remainder 4- 999999: 6 nines, remainder 0So yes, the first time we get a remainder of zero is with 6 nines. Therefore, k is 142857. But just to make sure I haven't missed any steps here.Alternatively, another approach is to model this as a congruence problem. We want 7k ≡ 0 mod 9...9 (with m digits). Wait, actually, if the number is composed of m nines, then it can be written as 10^m - 1. Because 999...9 (m times) is equal to 10^m - 1. For example, 999 = 10^3 - 1.So, we want 7k = 10^m - 1 for some natural number m. Therefore, 10^m ≡ 1 mod 7. So the problem reduces to finding the smallest m such that 10^m ≡ 1 mod 7, and then k would be (10^m - 1)/7. That's a better way to approach it, using modular arithmetic.So, m is the order of 10 modulo 7. The multiplicative order of 10 mod 7. Let's compute 10^m mod 7 and find the minimal m where this is 1.First, let's compute 10 mod 7: 10 ≡ 3 mod 7. So 10 ≡ 3 mod 7, so 10^m ≡ 3^m mod 7. So we need to find the minimal m such that 3^m ≡ 1 mod 7.Let's compute 3^m mod 7:- 3^1 = 3 mod 7- 3^2 = 9 ≡ 2 mod 7- 3^3 = 3*2 = 6 mod 7- 3^4 = 3*6 = 18 ≡ 4 mod 7- 3^5 = 3*4 = 12 ≡ 5 mod 7- 3^6 = 3*5 = 15 ≡ 1 mod 7Ah, so 3^6 ≡ 1 mod 7. Therefore, the multiplicative order of 3 mod 7 is 6, which means the multiplicative order of 10 mod 7 is also 6. Therefore, the smallest m is 6, so the minimal number of nines is 6, and k is (10^6 - 1)/7 = 999999/7 = 142857. So that's the answer.But let me check again if this is correct. 142857*7=999999. Let me do the multiplication:142857 * 7:7 * 100000 = 7000007 * 40000 = 280000 → 700000 + 280000 = 9800007 * 2000 = 14000 → 980000 + 14000 = 9940007 * 800 = 5600 → 994000 + 5600 = 9996007 * 50 = 350 → 999600 + 350 = 9999507 * 7 = 49 → 999950 + 49 = 999999Yes, that adds up. So 142857*7=999999, which is six nines. Therefore, 142857 is the smallest natural number that satisfies the condition.Just to confirm, since we know from modular arithmetic that the order is 6, there can't be a smaller m. Therefore, 6 is the minimal number of nines required, so 142857 is indeed the minimal k.Alternatively, if I tried to compute this by brute force, starting from k=1 and multiplying by 7, checking if the result is all nines. But that would take a long time. Instead, using modular arithmetic gives a much cleaner solution.Another way to think about it is that numbers consisting of all nines are of the form (10^m - 1)/9. But in this problem, we have 7k = (10^m - 1). So k = (10^m -1)/7. So we need (10^m -1) divisible by 7. Which again leads us to 10^m ≡1 mod 7. So the same conclusion.Therefore, the answer is 142857. So the smallest natural number is 142857.Final AnswerThe smallest natural number is boxed{142857}.</think>To find the smallest natural number ( k ) such that ( 7k ) is a number consisting only of nines, we need to determine the smallest ( m ) for which ( 10^m - 1 ) (a number with ( m ) nines) is divisible by 7. This translates to finding the smallest ( m ) such that ( 10^m equiv 1 pmod{7} ).First, we note that ( 10 equiv 3 pmod{7} ). Therefore, we need to find the smallest ( m ) such that ( 3^m equiv 1 pmod{7} ).Calculating powers of 3 modulo 7:- ( 3^1 equiv 3 pmod{7} )- ( 3^2 equiv 9 equiv 2 pmod{7} )- ( 3^3 equiv 3 cdot 2 = 6 pmod{7} )- ( 3^4 equiv 3 cdot 6 = 18 equiv 4 pmod{7} )- ( 3^5 equiv 3 cdot 4 = 12 equiv 5 pmod{7} )- ( 3^6 equiv 3 cdot 5 = 15 equiv 1 pmod{7} )The smallest ( m ) for which ( 3^m equiv 1 pmod{7} ) is 6. Therefore, ( 10^6 equiv 1 pmod{7} ), and the number consisting of six nines, ( 999999 ), is divisible by 7.Calculating ( k ) as ( frac{10^6 - 1}{7} ):[k = frac{999999}{7} = 142857]Thus, the smallest natural number ( k ) is (boxed{142857}).,<think>Okay, so I need to find the smallest natural number that when multiplied by 7 gives a number made up entirely of nines. Hmm, let's think about how to approach this.First, the problem is asking for a number composed only of nines. So, examples of such numbers would be 9, 99, 999, 9999, etc. Each of these is a string of nines with different lengths. So, essentially, I need to find the smallest number in this sequence (9, 99, 999, ...) that is divisible by 7. Then, the multiplier would be that number divided by 7.Another way to put it is: find the smallest k such that 999...9 (with k digits) is divisible by 7. Then, the answer is that number divided by 7. So, if I can find the smallest k where 10^k - 1 is divisible by 7, then the multiplier is (10^k - 1)/9 divided by 7. Wait, let me clarify.A number consisting of n nines is equal to (10^n - 1)/9 multiplied by 9, which is 10^n - 1. Wait, actually, a number with n nines is exactly equal to 10^n - 1 divided by 1, but wait, 10^1 - 1 = 9, 10^2 - 1 = 99, etc. So yes, a number with n nines is 10^n - 1. So, we need 10^n - 1 to be divisible by 7. Then, (10^n - 1)/7 would be the natural number we're looking for. So, the task reduces to finding the smallest n such that 10^n ≡ 1 mod 7. Because if 10^n ≡ 1 mod 7, then 10^n - 1 ≡ 0 mod 7, which means 7 divides (10^n - 1).So, this n is the multiplicative order of 10 modulo 7. The multiplicative order is the smallest positive integer n such that 10^n ≡ 1 mod 7. Let's compute that.First, compute 10 mod 7. 10 divided by 7 is 1 with a remainder of 3. So, 10 ≡ 3 mod 7. Therefore, 10^n ≡ 3^n mod 7. So, we need to find the smallest n where 3^n ≡ 1 mod 7.Let me compute powers of 3 modulo 7:3^1 = 3 mod 73^2 = 9 ≡ 2 mod 73^3 = 3*2 = 6 mod 73^4 = 3*6 = 18 ≡ 4 mod 73^5 = 3*4 = 12 ≡ 5 mod 73^6 = 3*5 = 15 ≡ 1 mod 7Ah, so 3^6 ≡ 1 mod 7. Therefore, the multiplicative order of 10 mod 7 is 6. Therefore, n = 6. Therefore, the smallest number composed of nines divisible by 7 is 999999 (which is 10^6 - 1). Then, the multiplier is 999999 / 7. Let's compute that.Dividing 999999 by 7:7*142857 = 999,999. Because 7*140,000 = 980,000. Then, 999,999 - 980,000 = 19,999. Then, 7*2,857 = 19,999 (since 7*2,000=14,000; 7*800=5,600; 14,000+5,600=19,600; 19,999-19,600=399; 7*57=399). So 142,857 + 2,857 = 145,714? Wait, wait, maybe my arithmetic is off.Wait, actually, I recall that 1/7 = 0.142857142857..., so 142857 * 7 = 999,999. Let me check that: 142,857 * 7.142,857 * 7:7 * 100,000 = 700,0007 * 40,000 = 280,0007 * 2,000 = 14,0007 * 800 = 5,6007 * 50 = 3507 * 7 = 49Add them all up: 700,000 + 280,000 = 980,000; 14,000 + 5,600 = 19,600; 350 + 49 = 399. Then, 980,000 + 19,600 = 999,600; 999,600 + 399 = 999,999. Yep, that works. So, 142,857 * 7 = 999,999.Therefore, the multiplier is 142,857, which when multiplied by 7 gives 999,999. So, 142,857 is the smallest such natural number.But let me verify this. Let's check smaller numbers. For example, 9 divided by 7 is 1.285..., which is not an integer. 99 divided by 7 is 14.142..., not an integer. 999 divided by 7 is 142.714..., not an integer. 9999 divided by 7 is 1428.428..., not an integer. 99999 divided by 7 is 14285.571..., not an integer. 999,999 divided by 7 is 142,857, which is an integer. So yes, the smallest n is 6, hence 142,857 is the answer.Alternatively, another way to approach this is to notice that we need 7x = 999...9. So, x = (999...9)/7. We need x to be an integer, so 999...9 must be divisible by 7. Let's try dividing 9 by 7: 9/7=1.285... Not integer. 99/7=14.142... No. 999/7=142.714... No. 9999/7=1428.428... No. 99999/7=14285.571... No. 999999/7=142857, which is integer. So, again, we get the same result.Therefore, the answer is 142857.But let me check if there's another way to think about this problem, maybe using modular arithmetic or something else. Since 10 ≡ 3 mod 7, as we saw before, so 10^n ≡ 3^n mod 7. So, we need 3^n ≡ 1 mod 7. We found that n=6 is the smallest such exponent. So, the length of the number consisting of nines is 6 digits. Therefore, the number is 999999, and dividing by 7 gives the multiplier 142857.Another thought: perhaps using the concept of repeating decimals. Since 1/7 has a repeating decimal of 6 digits: 0.142857142857..., which repeats every 6 digits. The repeating cycle of 1/7 is 142857, which is exactly the multiplier we found. So, this connects to the fact that the period of the decimal expansion of 1/7 is 6, which is the same as the multiplicative order of 10 modulo 7. Therefore, the length of the repeating cycle is the multiplicative order, which is 6. Hence, 10^6 -1 is divisible by 7, giving 999999, and 999999 /7 = 142857. Therefore, this confirms the answer again.Therefore, after multiple approaches and verifications, I can be confident that the smallest natural number to multiply 7 by to get a number consisting only of nines is 142857.Final AnswerThe smallest natural number is boxed{142857}.</think>To find the smallest natural number by which 7 must be multiplied to obtain a number consisting only of nines, we need to determine the smallest number composed entirely of nines that is divisible by 7. Such a number can be written as (10^n - 1) for some natural number (n). We need to find the smallest (n) such that (10^n equiv 1 pmod{7}).First, we note that (10 equiv 3 pmod{7}). Therefore, we need to find the smallest (n) such that (3^n equiv 1 pmod{7}). Calculating the powers of 3 modulo 7:[begin{align*}3^1 &equiv 3 pmod{7}, 3^2 &equiv 9 equiv 2 pmod{7}, 3^3 &equiv 3 cdot 2 = 6 pmod{7}, 3^4 &equiv 3 cdot 6 = 18 equiv 4 pmod{7}, 3^5 &equiv 3 cdot 4 = 12 equiv 5 pmod{7}, 3^6 &equiv 3 cdot 5 = 15 equiv 1 pmod{7}.end{align*}]The smallest (n) such that (3^n equiv 1 pmod{7}) is 6. Therefore, the smallest number consisting of nines that is divisible by 7 is (10^6 - 1 = 999999).Dividing this number by 7 gives:[999999 div 7 = 142857.]Thus, the smallest natural number by which 7 must be multiplied to obtain a number consisting only of nines is (boxed{142857}).]

🔑:triangle Let's find this number using divisiontext { 9999... } 7However, the quotient here is unknown, and the number of nines in the dividend is also unknown. We will perform the division by 7, appending nines to the dividend until the division is performed without a remainder for the first time. We get:begin{array}{ll}frac{999999}{29} & frac{7}{142857} frac{frac{19}{frac{59}{39}}}{frac{39}{frac{49}{0}}} & &end{array}Answer: 142857.