🧠:[<think>Okay, let's see. I have this problem here: Given that x = -1 is a solution to the equation x - 2m = 9, find m. Hmm, okay. So, they're telling me that when x is -1, the equation holds true. So, I need to find the value of m that makes this equation valid when x is -1. Alright, let me recall how to solve equations like this. The equation is linear, right? So, if x is known, then I can substitute that value into the equation and solve for m. That seems straightforward. Let me write down the equation again: x - 2m = 9. Since x is given as -1, substitute -1 in place of x. So, that would be -1 - 2m = 9. Now, I need to solve for m. Let me check if I substituted correctly. The original equation is x minus 2m equals 9. So replacing x with -1 gives me -1 minus 2m equals 9. Yes, that seems right. Now, the next step is to isolate m. So, first, maybe I can get the term with m by itself. Let's start by adding 1 to both sides of the equation to get rid of the -1 on the left. So, if I add 1 to both sides: -1 + 1 - 2m = 9 + 1. Which simplifies to 0 - 2m = 10. So, that's -2m = 10. Wait, hold on. Let me make sure. Left side: -1 - 2m + 1 would be (-1 + 1) which is 0, so -2m. Right side: 9 + 1 is 10. So, yes, that's correct. So, now the equation is -2m = 10. To solve for m, I need to divide both sides by -2. So, m = 10 / (-2) = -5. Hmm, so m is -5? Let me check that. Let me substitute m back into the original equation to verify. If m is -5, then the original equation becomes x - 2*(-5) = 9. Which simplifies to x + 10 = 9. Then, subtracting 10 from both sides gives x = 9 - 10 = -1. Which matches the given solution x = -1. So, that checks out. But let me go through my steps again just to be thorough. Starting with x - 2m = 9. Given x = -1. Substitute: -1 - 2m = 9. Then, add 1 to both sides: -2m = 10. Divide by -2: m = -5. Yep, that seems right. Alternatively, maybe I can approach it differently. Let's see. If I rearrange the original equation to solve for m first. Starting with x - 2m = 9. Subtract x from both sides: -2m = 9 - x. Then divide both sides by -2: m = (9 - x)/(-2). Then substitute x = -1 into this expression. So, m = (9 - (-1))/(-2) = (9 + 1)/(-2) = 10/(-2) = -5. Same answer. So, both methods give m as -5, which seems consistent. Another way to check is to plug both x and m back into the equation. If x is -1 and m is -5, then plugging in: (-1) - 2*(-5) = -1 + 10 = 9. Which is indeed equal to 9. Perfect. Wait, but just to be completely sure, maybe I made a mistake in the sign somewhere. Let me check the substitution again. Original equation: x - 2m = 9. If x is -1, then -1 -2m =9. So, solving for m: add 1 to both sides: -2m = 10. Divide by -2: m = -5. Signs: The term is -2m. So, when we divide both sides by -2, the equation becomes m = 10 / (-2) which is -5. Correct. Alternatively, if I had multiplied both sides by -1 first: -2m =10. Multiply by -1: 2m = -10. Then divide by 2: m = -5. Same result. Yes, I think that's solid. The answer is m = -5. But just to ensure, let's see if there's another perspective. For example, sometimes people might confuse coefficients or misapply substitution. Let me see. Suppose someone thought that m is part of the coefficient of x. But in the equation x - 2m =9, x is just x, and 2m is a separate term. So, substituting x with -1 is straightforward. Alternatively, if the equation were something like (x - 2)m =9, but in the problem it's written as x - 2m =9, so the structure is different. Parentheses can change things, but here they aren't present. So, it's x minus 2m equals 9, so substitution is clear. Alternatively, if someone misreads it as x - 2m =9 where x is a variable and m is a coefficient in another context, but the problem states that x = -1 is a solution, so substitution is the correct approach. Therefore, I think my answer is correct. m equals negative five.Final AnswerThe value of ( m ) is boxed{-5}.</think>Given that ( x = -1 ) is a solution to the equation ( x - 2m = 9 ), we need to find ( m ).1. Substitute ( x = -1 ) into the equation: [ -1 - 2m = 9 ]2. Add 1 to both sides of the equation to isolate the term with ( m ): [ -1 - 2m + 1 = 9 + 1 ] Simplifying this, we get: [ -2m = 10 ]3. Divide both sides by -2 to solve for ( m ): [ m = frac{10}{-2} = -5 ]4. Verification: Substitute ( x = -1 ) and ( m = -5 ) back into the original equation: [ -1 - 2(-5) = -1 + 10 = 9 ] This confirms the solution is correct.The value of ( m ) is (boxed{-5}).,<think>Okay, so I have this equation here: x minus 2m equals 9. And I'm told that x equals -1 is a solution. Hmm, let me think about what that means. If x is -1, then when I plug that into the equation, it should satisfy the equation, right? That means the left side should equal the right side once I substitute x with -1. Alright, let me write that out step by step. The original equation is x - 2m = 9. Since x = -1 is a solution, substituting x with -1 should make the equation true. So substituting gives me: (-1) - 2m = 9. Now, I need to solve for m here. Let me make sure I copied the problem correctly. The equation is x - 2m = 9, and x = -1 is a solution. Yeah, that seems right. So substituting x with -1 gives -1 - 2m = 9. Now, I need to isolate m. Let's see. First, maybe I can add 1 to both sides to get rid of the -1 on the left. So, adding 1 to both sides: (-1 - 2m) + 1 = 9 + 1. Simplifying the left side: -1 + 1 cancels out, leaving -2m. On the right side, 9 + 1 is 10. So now the equation is -2m = 10. Now, to solve for m, I need to divide both sides by -2. Let's do that: (-2m)/(-2) = 10/(-2). The left side simplifies to m, and the right side simplifies to -5. So m equals -5? Wait, let me check that again. If I substitute m with -5 back into the original equation with x = -1, does it hold? Let's check. Original equation: x - 2m = 9. Substitute x = -1 and m = -5: (-1) - 2*(-5) = -1 + 10 = 9. Yeah, that's correct. So m should be -5. But wait, let me go through the algebra again to make sure I didn't make a mistake. Starting with x = -1:Original equation: x - 2m = 9.Substitute x: -1 - 2m = 9.Add 1 to both sides: -2m = 10.Divide by -2: m = -5.Yes, that seems right. Alternatively, if I messed up the signs somewhere, could it be a different answer? Let me try m = 5. If m were 5, then substituting back: -1 -2*(5) = -1 -10 = -11, which isn't 9. So m can't be 5. What if I thought that adding 1 to both sides gave -2m = 8? Wait, no, 9 +1 is 10. So that step is correct. Then dividing 10 by -2 gives -5. Another way to check: Let's solve for m in terms of x first. The equation is x - 2m = 9. So, rearranged, we get -2m = 9 - x. Then m = (x - 9)/2. Wait, let me see: if we have -2m = 9 - x, then dividing both sides by -2 gives m = (9 - x)/(-2) = (x - 9)/2. So if x is -1, then m = (-1 - 9)/2 = (-10)/2 = -5. Yep, same result. That confirms it. So m is indeed -5. I think that's solid. No mistakes in the steps. The answer is m equals negative five.Final AnswerThe value of ( m ) is boxed{-5}.</think>Given that ( x = -1 ) is a solution to the equation ( x - 2m = 9 ), we substitute ( x = -1 ) into the equation:[-1 - 2m = 9]Next, we solve for ( m ). First, add 1 to both sides of the equation:[-1 - 2m + 1 = 9 + 1]This simplifies to:[-2m = 10]Then, divide both sides by -2:[m = frac{10}{-2} = -5]To verify, substitute ( m = -5 ) back into the original equation with ( x = -1 ):[-1 - 2(-5) = -1 + 10 = 9]This confirms the solution is correct. Therefore, the value of ( m ) is (boxed{-5}).]

🔑:To solve for m given that x=-1 is a solution to the equation x-2m=9, we substitute x=-1 into the equation:[-1 - 2m = 9]We then solve for m:[begin{align*}-1 - 2m &= 9 -2m &= 9 + 1 -2m &= 10 m &= frac{10}{-2} m &= -5end{align*}]Therefore, the value of m is boxed{-5}.

🧠:[<think>Okay, so I need to solve this geometry problem involving a parabola, its focus, directrix, and the area of triangle ABP. Let me start by recalling some basic properties of parabolas. A parabola is the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. The axis of symmetry is the line that passes through the focus and is perpendicular to the directrix. The vertex of the parabola is the midpoint between the focus and the directrix.The problem states that a line l passes through the focus of the parabola C and is perpendicular to the axis of symmetry. Since the axis of symmetry is perpendicular to the directrix, line l must be parallel to the directrix. Wait, actually, if the axis of symmetry is perpendicular to the directrix, then the directrix is a line perpendicular to the axis. Then, a line l that is perpendicular to the axis of symmetry would be parallel to the directrix. Hmm, maybe not. Let me think again.Actually, the axis of symmetry is a line that goes through the focus and is perpendicular to the directrix. So, if line l is perpendicular to the axis of symmetry, then line l must be parallel to the directrix. Because the directrix is also perpendicular to the axis of symmetry. So, line l is parallel to the directrix and passes through the focus. The line l intersects the parabola at points A and B, and the distance between A and B is 12. Then, P is a point on the directrix, and we need to find the area of triangle ABP. The answer choices are 18, 24, 36, or 48.Let me try to visualize this. Let's set up a coordinate system to make it easier. For a standard parabola, let's assume the axis of symmetry is the y-axis. Then the parabola can be written as y = (1/(4p))x², where p is the distance from the vertex to the focus (and also from the vertex to the directrix). The focus would be at (0, p), and the directrix is the line y = -p.Wait, but if the axis of symmetry is the y-axis, then the directrix is horizontal. Then, line l, which is perpendicular to the axis of symmetry (the y-axis), would be a horizontal line. But in the standard parabola, the directrix is horizontal as well, so line l is horizontal, same as the directrix. But the directrix is at y = -p, and line l passes through the focus at (0, p). So, line l is the horizontal line passing through (0, p), which is y = p.But wait, the line l intersects the parabola at points A and B. So, substituting y = p into the parabola equation y = (1/(4p))x², we get p = (1/(4p))x², so multiplying both sides by 4p gives 4p² = x², so x = ±2p. Therefore, points A and B are at (-2p, p) and (2p, p). The distance between A and B is the distance between these two points, which is 4p. But the problem states |AB| = 12, so 4p = 12, so p = 3.Therefore, the parabola is y = (1/(12))x², focus at (0, 3), directrix at y = -3. Then, the line l is y = 3, which intersects the parabola at (-6, 3) and (6, 3), so AB is 12 units long as given.Now, point P is any point on the directrix. The directrix is y = -3, so coordinates of P can be written as (h, -3), where h can be any real number. The problem asks for the area of triangle ABP. Since the area is the same regardless of where P is on the directrix (since the base AB is fixed and the height from P to AB is fixed), the area should be a constant value.Wait, is that true? Let me check. If P is on the directrix y = -3, then the height from P to the line AB (which is y = 3) is the vertical distance between y = 3 and y = -3, which is 6 units. The base AB is 12 units. Therefore, the area of triangle ABP is (1/2)*base*height = (1/2)*12*6 = 36. Therefore, the answer is C. 36.But let me confirm this reasoning. Since AB is the line segment between (-6, 3) and (6, 3), so the length is 12. The base is along y=3, and the height from any point on the directrix y=-3 to the line y=3 is 6 units. Therefore, regardless of the x-coordinate of P, the vertical distance remains 6. Thus, the area is (1/2)*12*6=36. So the answer is 36, which is option C.Alternatively, if I don't assume the standard parabola, but let me check if this holds for any parabola. Suppose the parabola is not necessarily oriented vertically. Wait, the problem says the line l is perpendicular to the axis of symmetry. So the axis of symmetry is some line, and line l is perpendicular to it. If the parabola is not standard, perhaps rotated, but the problem doesn't mention rotation, so likely it's a standard parabola with vertical or horizontal axis.But in the problem statement, the line l passes through the focus and is perpendicular to the axis of symmetry, so that would correspond to the latus rectum. Wait, the latus rectum is the line segment through the focus perpendicular to the axis of symmetry, with endpoints on the parabola. The length of the latus rectum is 4p for a standard parabola. Wait, in the standard parabola y²=4ax, the latus rectum is 4a. But in my previous calculation, for the parabola y = (1/(4p))x², which is equivalent to x²=4py, the latus rectum length is 4p. So when the line l (which is the latus rectum) intersects the parabola at A and B, |AB|=4p. The problem states |AB|=12, so 4p=12, so p=3. Then, the distance from the directrix to the line AB (which is the latus rectum) is from y=-3 to y=3, which is 6. Then, the area is (1/2)*12*6=36. So this seems consistent.Alternatively, if the parabola is horizontal, say x²=4py, but in that case, the axis of symmetry is the y-axis. Then, the focus is at (0,p), directrix is y=-p, line l is horizontal line through focus, y=p, which intersects parabola at x²=4p*p => x=±2p, so points (-2p, p) and (2p, p), length AB=4p. Then, same reasoning applies. So regardless of orientation, the result is same.Therefore, the area of triangle ABP is always 36, regardless of where P is on the directrix, since the height is fixed. Therefore, the answer should be C. 36.Final Answerboxed{C}</think>To solve the problem, we start by noting that the line ( l ) passes through the focus of the parabola ( C ) and is perpendicular to the axis of symmetry. This line ( l ) intersects the parabola at points ( A ) and ( B ) with ( |AB| = 12 ). We need to find the area of triangle ( triangle ABP ) where ( P ) is a point on the directrix of ( C ).1. Parabola Setup: - Assume the parabola is standard with the axis of symmetry as the y-axis. The equation is ( y = frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus (and the directrix is ( y = -p )). - The focus is at ( (0, p) ) and the directrix is ( y = -p ).2. Intersection of Line ( l ) with Parabola: - Line ( l ) is horizontal (perpendicular to the y-axis) and passes through the focus, so its equation is ( y = p ). - Substituting ( y = p ) into the parabola equation ( y = frac{1}{4p}x^2 ), we solve for ( x ): [ p = frac{1}{4p}x^2 implies 4p^2 = x^2 implies x = pm 2p ] - Points ( A ) and ( B ) are ( (-2p, p) ) and ( (2p, p) ). The distance ( |AB| = 4p ).3. Determine ( p ): - Given ( |AB| = 12 ), we find ( 4p = 12 implies p = 3 ).4. Area Calculation: - The directrix is ( y = -3 ). Any point ( P ) on the directrix has coordinates ( (h, -3) ). - The height from ( P ) to the line ( AB ) (which is ( y = 3 )) is the vertical distance ( 3 - (-3) = 6 ). - The base ( AB ) is 12. The area of ( triangle ABP ) is: [ text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 12 times 6 = 36 ]Thus, the area of triangle ( triangle ABP ) is (boxed{C}).,<think>Okay, let's see. I need to find the area of triangle ABP where P is a point on the directrix of parabola C. The line l passes through the focus and is perpendicular to the axis of symmetry, intersecting the parabola at A and B with |AB| = 12. The options are 18, 24, 36, 48. Hmm.First, let me recall some properties of parabolas. The standard form of a parabola with vertical axis of symmetry is y = (1/(4p))x², where p is the distance from the vertex to the focus (and also from the vertex to the directrix). The focus is at (0, p) and the directrix is the line y = -p. If the axis of symmetry is vertical, then a line perpendicular to it would be horizontal. So line l is horizontal, passes through the focus, which is (0, p) if the parabola is oriented upwards. So the equation of line l would be y = p.This line intersects the parabola at points A and B. Let's find those points. Substitute y = p into the parabola equation:p = (1/(4p))x²Multiply both sides by 4p:4p² = x²So x = ±2pTherefore, points A and B are at (-2p, p) and (2p, p), so the distance between them is |2p - (-2p)| = 4p. But the problem states that |AB| = 12, so 4p = 12 => p = 3. So the distance from vertex to focus is 3, and the directrix is y = -3 if the parabola opens upwards. Wait, but the problem doesn't specify the orientation of the parabola. Hmm. Wait, the axis of symmetry is the axis perpendicular to line l, which is horizontal, so the axis of symmetry must be vertical. Therefore, the parabola opens either upwards or downwards. Let's assume it's upwards for simplicity.So the parabola is y = (1/(12))x², since 4p = 12? Wait, wait, no. Wait, the standard form is y = (1/(4p))x². Since p = 3, then 4p = 12, so the equation is y = (1/12)x². The focus is at (0, 3) and the directrix is y = -3.Now, point P is on the directrix. The directrix is y = -3, so any point P on the directrix has coordinates (h, -3) where h can be any real number.We need to find the area of triangle ABP. The coordinates of A and B are (-6, 3) and (6, 3) since p = 3, so 2p = 6. So AB is a horizontal line segment from (-6, 3) to (6, 3), length 12, which matches.Point P is (h, -3). The area of triangle ABP can be calculated using the formula for the area of a triangle given three vertices. The coordinates are A(-6, 3), B(6, 3), P(h, -3).The formula is |(Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By))/2|. Wait, but let's label the points correctly. Let me use coordinates (x1, y1), (x2, y2), (x3, y3). So A(-6, 3), B(6, 3), P(h, -3). Plugging into the area formula:Area = |( (-6)(3 - (-3)) + 6((-3) - 3) + h(3 - 3) ) / 2 |Simplify each term:First term: (-6)(3 - (-3)) = (-6)(6) = -36Second term: 6((-3) - 3) = 6(-6) = -36Third term: h(3 - 3) = h(0) = 0So total inside the absolute value: (-36 -36 + 0)/2 = (-72)/2 = -36. Taking absolute value, 36.Wait, so the area is 36 regardless of h? That's interesting. So no matter where point P is on the directrix, the area of triangle ABP is always 36? Let me check that again.Alternatively, maybe I can use another method. Since AB is a horizontal line from (-6,3) to (6,3), its length is 12. The base of the triangle can be AB, which is 12 units. The height would be the vertical distance from point P to the line AB. Since AB is at y = 3, and point P is on y = -3, the vertical distance is |3 - (-3)| = 6. So the height is 6. Then the area is (base * height)/2 = (12 * 6)/2 = 72/2 = 36. So yes, that's another way. Therefore, regardless of where P is on the directrix, the vertical distance is always 6, so the area is always 36.Therefore, the answer should be C. 36.Final Answerboxed{C}</think>Given a parabola ( C ) with its axis of symmetry along the vertical axis, the line ( l ) passes through the focus and is perpendicular to the axis of symmetry, intersecting the parabola at points ( A ) and ( B ) such that ( |AB| = 12 ). We need to find the area of triangle ( triangle ABP ) where ( P ) is a point on the directrix of ( C ).1. Identify the parabola properties: - The standard form of the parabola is ( y = frac{1}{4p}x^2 ) where ( p ) is the distance from the vertex to the focus. - The line ( l ) is horizontal (perpendicular to the vertical axis of symmetry) and passes through the focus ( (0, p) ), so its equation is ( y = p ).2. Find the points of intersection ( A ) and ( B ): - Substitute ( y = p ) into the parabola equation: ( p = frac{1}{4p}x^2 ). - Solving for ( x ), we get ( x = pm 2p ). Thus, points ( A ) and ( B ) are ( (-2p, p) ) and ( (2p, p) ). - The distance ( |AB| = 4p ). Given ( |AB| = 12 ), we find ( 4p = 12 ) hence ( p = 3 ).3. Determine the coordinates: - The parabola equation is ( y = frac{1}{12}x^2 ). - The focus is at ( (0, 3) ) and the directrix is ( y = -3 ). - Points ( A ) and ( B ) are ( (-6, 3) ) and ( (6, 3) ).4. Calculate the area of ( triangle ABP ): - Point ( P ) on the directrix has coordinates ( (h, -3) ). - The base ( AB ) is 12 units. The height is the vertical distance from ( P ) to ( AB ), which is ( |3 - (-3)| = 6 ). - The area is ( frac{1}{2} times text{base} times text{height} = frac{1}{2} times 12 times 6 = 36 ).Thus, the area of ( triangle ABP ) is (boxed{C}).]

🔑:Since the line l passes through the focus of the parabola and is perpendicular to its axis of symmetry, the segment AB is a chord of the parabola that passes through its focus. According to the properties of a parabola, the perpendicular bisector of any chord passing through the focus also passes through the corresponding point on the directrix. Therefore, the distance from the midpoint of AB to the directrix is equal to half the length of AB, which is frac{12}{2}=6. The area of triangle ABP can be calculated as half the product of the base AB and the height (the distance from the midpoint of AB to the directrix), which is frac{1}{2} times 12 times 6 = 36.Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, let's try to figure out this quadratic function problem. The question says that the vertex of the quadratic function y = ax² + bx + c is on the negative half-axis of the y-axis, and the part to the left of its axis of symmetry is rising. We need to write an analytical expression for such a quadratic function. Hmm, let me break this down step by step.First, let's recall some basics about quadratic functions. A quadratic function in standard form is y = ax² + bx + c. The vertex of this parabola is at the point (h, k), where h = -b/(2a) and k is the value of the function at x = h, so k = c - b²/(4a). The axis of symmetry is the vertical line x = h. The parabola opens upwards if a > 0 and downwards if a < 0.Now, the problem states that the vertex is on the negative half-axis of the y-axis. That means the vertex has coordinates (0, k) where k < 0. Wait, because the y-axis is the line x = 0, so if the vertex is on the negative half-axis of the y-axis, its x-coordinate must be 0, and its y-coordinate is negative. Therefore, the vertex is at (0, k) with k < 0. So, substituting h = 0 into the vertex formula, since h is the x-coordinate of the vertex. But h = -b/(2a) = 0. Therefore, -b/(2a) = 0 implies that b = 0. So, the quadratic function simplifies to y = ax² + c. That's because the b term is zero.Moreover, the vertex is at (0, c), because in the standard form y = ax² + bx + c, if b = 0, then the vertex is indeed (0, c). Since the vertex is on the negative y-axis, c must be less than 0. So, c < 0.Next, the problem mentions that the part to the left of its axis of symmetry is rising. The axis of symmetry is x = 0 in this case because the vertex is at (0, c). To the left of the axis of symmetry means for x < 0. The function is rising (increasing) as we move from left to right in that region. Now, since the axis of symmetry is x = 0, the left side is x < 0. Let's recall that for a quadratic function, if a > 0, the parabola opens upwards, which means it decreases to the left of the vertex and increases to the right. Conversely, if a < 0, the parabola opens downwards, increasing to the left of the vertex and decreasing to the right.Wait, so if the part to the left of the axis of symmetry (x < 0) is rising, that would mean that as x increases from -∞ to 0 (moving left to right), the function's value increases. If the parabola opens upwards (a > 0), then to the left of the vertex (x < 0), the function is decreasing. But if it opens downward (a < 0), then to the left of the vertex, the function is increasing. Therefore, to have the left side rising, the parabola must open downward, so a < 0.Putting this all together: the quadratic function must be of the form y = ax² + c with a < 0 and c < 0. So, for example, we can choose specific values for a and c that satisfy these inequalities.Let me verify this. Let's take a simple example: y = -x² - 1. Here, a = -1 (which is less than 0) and c = -1 (which is less than 0). The vertex is at (0, -1), which is on the negative y-axis. Since a is negative, the parabola opens downward. Therefore, to the left of the axis of symmetry (x < 0), the function is increasing. Let me check the derivative to be sure. The derivative of y = -x² -1 is dy/dx = -2x. For x < 0, the derivative is positive (since -2x would be positive when x is negative), so the function is indeed increasing on x < 0. That fits the conditions.Alternatively, another example could be y = -2x² - 3. Here, a = -2 and c = -3. The vertex is at (0, -3), which is on the negative y-axis. The parabola opens downward because a is negative. The derivative is dy/dx = -4x. For x < 0, dy/dx is positive, so increasing. That also works.Therefore, any quadratic function of the form y = ax² + c with a < 0 and c < 0 will satisfy the given conditions. The problem just asks to write one such example. So, the simplest one is probably y = -x² -1. Alternatively, even y = -x² - 2 would work, but maybe the first is the most straightforward.Wait, but let me confirm once again. The vertex is at (0, c). Since c is negative, that's on the negative y-axis. The parabola opens downward (a < 0), so to the left of x = 0, which is x < 0, the function is increasing. The right side would be decreasing. So, yes, that's exactly what the problem states. So, this should be correct.Alternatively, if I had chosen a different form where b is not zero, but the problem states the vertex is on the y-axis, so x-coordinate of the vertex is 0, which forces b = 0. Therefore, the quadratic must be of the form y = ax² + c. So, any answer must have b = 0, and a and c as negative numbers.Therefore, one possible analytical expression is y = -x² -1. Let me just check again. The vertex is (0, -1), which is on the negative y-axis. The axis of symmetry is x = 0. The coefficient a is -1, so the parabola opens downward. Therefore, to the left of the axis of symmetry (x < 0), the function is increasing. Yes, because as x increases from -∞ to 0, the function's value goes from -∞ up to -1. So, that is rising. On the right side (x > 0), the function decreases from -1 to -∞. So, that's correct.Alternatively, if I had chosen y = -2x² - 5, that would also work. The problem says "just write one," so the simplest one is probably best.But let me check if there's any other consideration. For example, does the quadratic have to have real roots? The problem doesn't specify, so it's okay if it doesn't. Since c is negative and a is negative, the quadratic equation ax² + c = 0 would have solutions x = ±sqrt(-c/a). Since both a and c are negative, -c/a is positive, so sqrt of a positive number is real. Therefore, the quadratic would have two real roots. But again, the problem doesn't mention anything about roots, so that's irrelevant. So, even if it didn't have real roots, it would still be okay. But in this case, with a and c negative, it does have real roots. But that's beside the point.Therefore, the answer can be any quadratic of the form y = ax² + c where a is negative and c is negative. So, writing one such example: y = -x² - 1. That's a valid answer. Alternatively, even y = -x² would not work because the vertex is at (0, 0), which is on the origin, not the negative y-axis. So c has to be strictly less than 0. So, y = -x² - 1 is good.So, to recap: the vertex is on the negative y-axis, so (0, c) with c < 0. The function rises to the left of the axis of symmetry (x=0), which requires the parabola to open downward, so a < 0. Therefore, the quadratic is y = ax² + c with a < 0 and c < 0. Example: y = -x² -1.I think that's solid. Let me just check if there's a different interpretation of the problem. The axis of symmetry's left part is rising. If the axis of symmetry wasn't the y-axis, but the vertex was on the y-axis, but the axis of symmetry is still the y-axis. So, in all cases, the axis of symmetry is x = 0 here. So, yes, left of x=0 is rising. So, that's consistent. Therefore, the example is correct.Final AnswerThe analytical expression of the quadratic function can be boxed{y = -x^2 - 1}.</think>The vertex of the quadratic function ( y = ax^2 + bx + c ) is on the negative half-axis of the ( y )-axis, meaning the vertex coordinates are ( (0, k) ) with ( k < 0 ). This implies ( b = 0 ) and the quadratic function simplifies to ( y = ax^2 + c ) with ( c < 0 ).Since the part to the left of the axis of symmetry (which is the ( y )-axis, ( x = 0 )) is rising, the parabola must open downwards. This requires ( a < 0 ).An example of such a quadratic function is ( y = -x^2 - 1 ). Here, ( a = -1 ) (which is negative) and ( c = -1 ) (which is negative). The vertex is at ( (0, -1) ), on the negative ( y )-axis, and the parabola opens downward, making the function increase to the left of the ( y )-axis.[boxed{y = -x^2 - 1}],<think>Okay, so I need to find a quadratic function in the form y = ax² + bx + c where the vertex is on the negative half of the y-axis, and the part to the left of its axis of symmetry is rising. Hmm, let me break this down step by step.First, the vertex of a quadratic function is at the point (-b/(2a), c - b²/(4a)). Wait, actually, let me recall the formula correctly. The vertex (h, k) can be found using h = -b/(2a) and k = f(h) = c - b²/(4a). Yeah, that's right. So the y-coordinate of the vertex is k = c - b²/(4a), and the problem states that this vertex is on the negative half of the y-axis. That means the vertex is at (h, k) where k is negative. Since it's on the negative half-axis of the y-axis, the x-coordinate of the vertex must be zero? Wait, hold on. The negative half-axis of the y-axis is the part of the y-axis where y is negative. So the vertex lies somewhere along the y-axis below the origin. So that would mean the vertex is at (0, k) where k < 0. Because if the vertex is on the y-axis, its x-coordinate is zero. So h = 0. Therefore, from h = -b/(2a) = 0, this implies that -b/(2a) = 0, so b must be zero. Okay, so b = 0. Then the quadratic simplifies to y = ax² + c. The vertex is at (0, c), since if b = 0, then k = c - 0 = c. Therefore, c must be negative. So c < 0.Next part: "the part to the left of its axis of symmetry is rising." The axis of symmetry of a quadratic function is the vertical line x = h, which in this case is x = 0, since the vertex is at (0, c). So the axis of symmetry is the y-axis itself. The part to the left of the axis of symmetry (x < 0) is rising, which means that as x increases from negative values towards 0, the function increases. So the function is increasing on the interval (-∞, 0) if the parabola opens upwards, but wait, if a parabola opens upwards, it decreases to the left of the vertex and increases to the right. Wait, no. Wait, the standard parabola y = ax² + c with a > 0 opens upwards, so it decreases as you move left from the vertex (since the vertex is the minimum point). If a < 0, it opens downward, so to the left of the vertex, it would be increasing.Wait, let me think again. For a quadratic function y = ax² + bx + c, if a > 0, it opens upwards, so the vertex is a minimum point. Therefore, to the left of the axis of symmetry (x = h), the function is decreasing, and to the right, it's increasing. Wait, no. Wait, suppose the vertex is at (h, k). If a > 0, then as you move from left to right, the function decreases until it reaches the vertex, then increases. So for x < h, the function is decreasing, and for x > h, it's increasing. So if the axis of symmetry is x = 0, then for x < 0 (left of the axis), the function is decreasing, and for x > 0, increasing. But the problem says that the part to the left of the axis of symmetry is rising. So it's increasing as x increases. Therefore, the function must be increasing for x < 0. But if a > 0, the function is decreasing for x < h. If a < 0, then the parabola opens downward, so the vertex is a maximum. Then, for x < h, the function is increasing, and for x > h, it's decreasing. So if a < 0, then to the left of the axis of symmetry (x = 0), which is x < 0, the function is increasing. That's exactly what we need. So a must be negative.So combining this, since the vertex is at (0, c) with c < 0, and the parabola opens downward (a < 0), then such a quadratic function would satisfy the given conditions.Therefore, the quadratic function can be written as y = -x² + c, where c is a negative constant. For simplicity, let's choose c = -1. Then the function would be y = -x² - 1. Let's check the conditions.Vertex is at (0, -1), which is on the negative y-axis. The parabola opens downward (a = -1 < 0), so to the left of the axis of symmetry (x < 0), the function is increasing. Yes, because as x increases from -∞ to 0, the y-values go from -∞ up to -1. So it is indeed rising to the left of the axis.Alternatively, another example could be y = -2x² - 3, as long as a is negative and c is negative. But the problem just asks to write one example, so the simplest one is probably y = -x² - 1.Wait, but the original problem says "the analytical expression of this quadratic function can be ______ (just write one)." So I can choose any quadratic function that meets the conditions. Let me verify once more.Given y = ax² + bx + c. Since the vertex is on the negative y-axis, b must be zero (since h = -b/(2a) = 0 implies b = 0). Then c must be negative. Then the coefficient a must be negative so that the parabola opens downward, making the function increase to the left of the y-axis. Therefore, the general form is y = ax² + c where a < 0 and c < 0. So examples include y = -x² - 1, y = -2x² - 5, etc. The simplest one is y = -x² - 1. Let me confirm that with the conditions:- Vertex is (0, -1), which is on the negative y-axis. Check.- The coefficient a is -1, which is negative, so the parabola opens downward. Therefore, to the left of the axis of symmetry (x < 0), the function is increasing. For example, when x = -2, y = -(-2)² -1 = -4 -1 = -5. When x = -1, y = -1 -1 = -2. When x approaches 0 from the left, y approaches -1. So as x increases from -∞ to 0, y increases from -∞ to -1. Hence, it's rising to the left of the axis. Check.Therefore, the answer is y = -x² - 1. Alternatively, another valid answer would be y = -3x² - 2, but since the problem asks for just one, the simplest is y = -x² - 1.Final AnswerThe analytical expression of the quadratic function can be boxed{y = -x^2 - 1}.</think>The vertex of the quadratic function ( y = ax^2 + bx + c ) is on the negative half-axis of the ( y )-axis, implying the vertex is at ((0, k)) where ( k < 0 ). This means ( b = 0 ) (since the x-coordinate of the vertex is ( -frac{b}{2a} = 0 )) and ( c = k ) (since the y-coordinate of the vertex is ( c - frac{b^2}{4a} = c )). Therefore, the quadratic function simplifies to ( y = ax^2 + c ) with ( c < 0 ).Additionally, the part to the left of the axis of symmetry (the ( y )-axis) is rising. For the left side of the parabola to be rising, the parabola must open downward, which means ( a < 0 ).A simple example of such a quadratic function is ( y = -x^2 - 1 ). Here, the vertex is at ((0, -1)), which is on the negative ( y )-axis, and the coefficient ( a = -1 ) ensures the parabola opens downward, making the function rise to the left of the ( y )-axis.Thus, the analytical expression of the quadratic function can be (boxed{y = -x^2 - 1}).]

🔑:To find the analytical expression of the quadratic function y=ax^{2}+bx+c that meets the given conditions, we proceed as follows:1. Condition for the vertex to be on the negative half-axis of the y-axis: The vertex's y-coordinate is given by c when b=0. For the vertex to be on the negative half-axis, c must be negative. Thus, we have c < 0.2. Condition for the part to the left of its axis of symmetry to be rising: This implies that the parabola opens downwards, which happens when a < 0.3. Condition for the axis of symmetry to be vertical (not slanted): This is achieved when b=0, ensuring the parabola is symmetric about the y-axis.Given these conditions, we can construct a quadratic function that satisfies all: b=0, a < 0, and c < 0. A simple example that meets these criteria is y=-x^{2}-1. Therefore, the analytical expression of this quadratic function can be written as:[y=-x^{2}-1]Hence, the answer is:[boxed{y=-x^{2}-1}]Note: The answer is not unique, and any quadratic function that satisfies the given conditions is acceptable.

🧠:[<think>Okay, let's see. I need to find the length of CD on a semicircle with diameter AD = 3. Points B and C are placed such that AB = BC = 1. Hmm, so the semicircle has diameter AD, which is 3 units long. That means the radius of the semicircle is 1.5 units. Let me visualize this: points A and D are at the ends of the diameter, and B and C are two points on the semicircle between A and D. The distances from A to B and from B to C are both 1 unit. I need to find the length from C to D.First, maybe I can place this semicircle on a coordinate system to make it easier. Let me set the center of the semicircle at the origin (0,0), so that point A is at (-1.5, 0) and point D is at (1.5, 0). The semicircle is the upper half of the circle with equation x² + y² = (1.5)² = 2.25.Points B and C are on the semicircle such that AB = 1 and BC = 1. So starting from A, moving 1 unit along the semicircle to B, then another 1 unit to C, and then from C to D is the remaining distance. But wait, these distances AB and BC are chord lengths, not arc lengths, right? Because the problem says "two points B and C are placed such that AB = BC = 1". So AB and BC are straight-line distances, not along the arc.Therefore, points A, B, C, D are all on the semicircle with AD as diameter. The chord length from A to B is 1, from B to C is 1, and we need the chord length from C to D.Given that the diameter AD is 3, which is the longest chord in the semicircle, and the radius is 1.5. Let me recall that the chord length between two points on a circle is given by 2r sin(θ/2), where θ is the central angle subtended by the chord, and r is the radius.Since AB = 1, the chord length AB is 1. So, 2 * 1.5 * sin(θ_AB / 2) = 1. That would be 3 sin(θ_AB / 2) = 1, so sin(θ_AB / 2) = 1/3. Therefore, θ_AB / 2 = arcsin(1/3), so θ_ab = 2 arcsin(1/3). Similarly, chord BC = 1, so the central angle θ_bc is also 2 arcsin(1/3). Therefore, the central angles between A and B, and between B and C, are each 2 arcsin(1/3). Then, the central angle between C and D would be the remaining angle in the semicircle.Since the semicircle is 180 degrees, or π radians. So, total central angle from A to D is π. If we have θ_ab + θ_bc + θ_cd = π. So θ_cd = π - 2 * 2 arcsin(1/3) = π - 4 arcsin(1/3). Then, the chord length CD is 2 * 1.5 * sin(θ_cd / 2) = 3 sin( (π - 4 arcsin(1/3)) / 2 ).But maybe there is a better way to approach this. Alternatively, using coordinates. Let me try coordinates.Let me set up the coordinate system: center at O(0,0), radius 1.5. A is (-1.5,0), D is (1.5,0). Points B and C are on the semicircle (upper half). Let me denote coordinates of B as (x1, y1) and C as (x2, y2). Since AB = 1, the distance from A(-1.5,0) to B(x1,y1) is 1. So:√[(x1 + 1.5)^2 + (y1 - 0)^2] = 1.Similarly, distance from B(x1,y1) to C(x2,y2) is 1:√[(x2 - x1)^2 + (y2 - y1)^2] = 1.And both B and C lie on the semicircle x² + y² = 2.25, with y ≥ 0.So we have three equations:1. (x1 + 1.5)^2 + y1^2 = 1.2. (x2 - x1)^2 + (y2 - y1)^2 = 1.3. x1^2 + y1^2 = 2.25.4. x2^2 + y2^2 = 2.25.So equations 1 and 3 can be used to find coordinates of B. Similarly, equations 2, 3, and 4 can be used for coordinates of C.Let me first solve for point B.From equation 1: (x1 + 1.5)^2 + y1^2 = 1.From equation 3: x1^2 + y1^2 = 2.25.Subtract equation 3 from equation 1:(x1 + 1.5)^2 - x1^2 = 1 - 2.25 = -1.25.Expanding the left side: x1² + 3x1 + 2.25 - x1² = 3x1 + 2.25.Therefore, 3x1 + 2.25 = -1.25.So, 3x1 = -1.25 - 2.25 = -3.5.Therefore, x1 = -3.5 / 3 ≈ -1.1667.But wait, AD is from -1.5 to 1.5 on the x-axis, so x1 should be between -1.5 and 1.5. Let's compute x1 = -3.5 / 3 ≈ -1.166666..., which is between -1.5 and 1.5, so that's okay.Then, from equation 3: x1^2 + y1^2 = 2.25, so y1 = sqrt(2.25 - x1^2).Compute x1^2: (-3.5/3)^2 = (12.25)/9 ≈ 1.3611.Thus, y1^2 = 2.25 - 1.3611 ≈ 0.8889, so y1 ≈ sqrt(0.8889) ≈ 0.9428.Since it's the upper semicircle, y1 is positive.So coordinates of B are approximately (-1.1667, 0.9428).Now, moving on to point C. We know that the distance from B to C is 1, and C is on the semicircle. Let's denote C as (x2, y2). Then:(x2 - x1)^2 + (y2 - y1)^2 = 1.And x2² + y2² = 2.25.So we have two equations with variables x2 and y2.This seems a bit complex. Maybe we can use coordinates with variables.Let me denote x1 = -3.5/3 = -7/6 ≈ -1.1667. Wait, -3.5/3 is -7/6, right? Because 3.5 is 7/2, so 7/2 divided by 3 is 7/6. So x1 = -7/6. Then y1^2 = 2.25 - (49/36) = (81/36 - 49/36) = 32/36 = 8/9. So y1 = 2√2 / 3 ≈ 0.9428. That's exact value.So coordinates of B are (-7/6, (2√2)/3).Now, let's denote point C as (x2, y2). Then:(x2 + 7/6)^2 + (y2 - (2√2)/3)^2 = 1. [Wait, distance from B to C: x2 - x1 = x2 - (-7/6) = x2 + 7/6, and y2 - y1 = y2 - (2√2)/3. So yes.]Also, x2² + y2² = 9/4 = 2.25, since radius is 3/2.So we have:Equation 1: (x2 + 7/6)^2 + (y2 - (2√2)/3)^2 = 1.Equation 2: x2² + y2² = 9/4.Let me expand equation 1:(x2 + 7/6)^2 + (y2 - (2√2)/3)^2 = x2² + (7/3)x2 + 49/36 + y2² - (4√2)/3 y2 + 8/9 = 1.So combining terms:x2² + y2² + (7/3)x2 - (4√2)/3 y2 + 49/36 + 8/9 - 1 = 0.From equation 2, x2² + y2² = 9/4. Substitute that into equation 1:9/4 + (7/3)x2 - (4√2)/3 y2 + 49/36 + 8/9 - 1 = 0.Convert all constants to 36 denominators:9/4 = 81/36,49/36 remains,8/9 = 32/36,1 = 36/36.So:81/36 + (7/3)x2 - (4√2)/3 y2 + 49/36 + 32/36 - 36/36 = 0.Compute constants:81 + 49 + 32 - 36 = 81 + 49 = 130; 130 + 32 = 162; 162 -36=126. So 126/36.Therefore:126/36 + (7/3)x2 - (4√2)/3 y2 = 0.Simplify 126/36 = 7/2.So:7/2 + (7/3)x2 - (4√2)/3 y2 = 0.Multiply both sides by 6 to eliminate denominators:21 + 14x2 - 8√2 y2 = 0.Thus:14x2 - 8√2 y2 = -21.So equation 3: 14x2 - 8√2 y2 = -21.Now, equation 2: x2² + y2² = 9/4.So we have two equations: equation 2 and equation 3. Let's solve for x2 and y2.From equation 3: 14x2 - 8√2 y2 = -21.Let me solve for one variable in terms of the other. Let's solve for x2:14x2 = 8√2 y2 -21.So x2 = (8√2 y2 -21)/14.Now substitute into equation 2:[(8√2 y2 -21)/14]^2 + y2² = 9/4.This will be a quadratic in y2, but seems complicated. Let me compute each term step by step.First, compute [(8√2 y2 -21)/14]^2:Let me denote numerator as (8√2 y2 -21). Then squared:(8√2 y2 -21)^2 = (8√2 y2)^2 - 2*8√2 y2*21 + 21^2 = 128 y2² - 336√2 y2 + 441.So, divided by 14² = 196:[128 y2² - 336√2 y2 + 441]/196.Thus, equation 2 becomes:[128 y2² - 336√2 y2 + 441]/196 + y2² = 9/4.Multiply all terms by 196 to eliminate denominators:128 y2² - 336√2 y2 + 441 + 196 y2² = 9/4 *196 = 441.So:(128 + 196) y2² - 336√2 y2 + 441 + 441 - 441 = 0.Wait, wait. Let's do it step by step.Left side:[128 y2² - 336√2 y2 + 441] + 196 y2² = 9/4 *196.So:128 y2² + 196 y2² - 336√2 y2 + 441 = (9/4)*196.Compute 128 + 196 = 324.So:324 y2² - 336√2 y2 + 441 = 441.Since (9/4)*196 = 9*49 = 441. Yes, right side is 441.Therefore:324 y2² - 336√2 y2 + 441 = 441.Subtract 441 from both sides:324 y2² - 336√2 y2 = 0.Factor out y2:y2 (324 y2 - 336√2 ) = 0.Thus, solutions are:y2 = 0, or 324 y2 - 336√2 = 0 => y2 = (336√2)/324 = simplify.Simplify 336/324: divide numerator and denominator by 12: 28/27. So y2 = (28√2)/27 ≈ 28*1.4142 /27 ≈ 1.456.But since point C is on the upper semicircle, y2 > 0. If y2 = 0, then point C would be on the diameter, but since C is on the semicircle, y2 =0 is possible only if C is A or D. But given that BC =1, and B is at y ≈0.9428, C can't be A or D. Therefore, the valid solution is y2 = (28√2)/27.Then, substitute back into equation 3: 14x2 -8√2 y2 = -21.So x2 = (8√2 y2 -21)/14.Plug in y2 = (28√2)/27:x2 = (8√2*(28√2)/27 -21)/14.Compute 8√2*(28√2) = 8*28*(√2)^2 = 224*2 = 448.Thus:x2 = (448/27 -21)/14.Convert 21 to 567/27:x2 = (448/27 - 567/27)/14 = (-119/27)/14 = -119/(27*14) = -119/378 = -17/54 ≈ -0.3148.Wait, but let me check the arithmetic:448 - 567 = -119. So yes, numerator is -119/27.Divide by 14: (-119/27)/14 = -119/(27*14). 14 divides into 119: 119 ÷14 = 8.5, but let's see prime factors:119 is 7*17. 14 is 2*7. So 119/14 = 17/2. But since it's negative:-119/(27*14) = -17/(27*2) = -17/54.Yes, because 119 = 7*17, 14 =7*2, so cancel 7: 17/ (27*2) =17/54. So x2 = -17/54.Therefore, coordinates of point C are (-17/54, 28√2/27).Now, we need to find CD, the distance from C(-17/54, 28√2/27) to D(3/2, 0).Compute the distance:√[(3/2 - (-17/54))² + (0 - 28√2/27)²].First, compute the x-coordinate difference:3/2 + 17/54. Convert 3/2 to 81/54:81/54 +17/54 = 98/54 = 49/27.Y-coordinate difference is -28√2/27.Therefore, CD = √[(49/27)^2 + ( -28√2/27)^2] = √[(2401/729) + (784*2)/729] = √[(2401 + 1568)/729] = √[3969/729] = √[3969]/√729 = 63/27 = 7/3 ≈ 2.3333.Wait, 63 divided by 27 is 7/3. So CD = 7/3.But let me verify the calculation:First, (49/27)^2 = 49² /27² = 2401 /729.(28√2/27)^2 = (28² * 2)/27² = 784 * 2 /729 = 1568 /729.Adding 2401 +1568 = 3969. 3969 /729 = 5.444... But 3969 divided by 729: 729*5=3645, 3969-3645=324. 324/729= 4/9. So total is 5 + 4/9 = 49/9. Therefore, sqrt(49/9)=7/3.Yes, so CD=7/3.Therefore, the length CD is 7/3, which is approximately 2.333...So the answer should be boxed{dfrac{7}{3}}.Final AnswerThe length ( CD ) is boxed{dfrac{7}{3}}.</think>Given a semicircle with diameter (AD = 3) and radius (1.5), points (B) and (C) are placed such that (AB = BC = 1). We need to find the length (CD).1. Setting up the coordinate system: - The center of the semicircle is at the origin ((0,0)). - Points (A) and (D) are at ((-1.5, 0)) and ((1.5, 0)) respectively. - The equation of the semicircle is (x^2 + y^2 = 2.25) with (y geq 0).2. Finding coordinates of point (B): - The distance (AB = 1) gives the equation: (sqrt{(x_1 + 1.5)^2 + y_1^2} = 1). - Solving this with the semicircle equation (x_1^2 + y_1^2 = 2.25): - Subtracting the equations: ((x_1 + 1.5)^2 - x_1^2 = -1.25). - Solving gives (x_1 = -frac{7}{6}) and (y_1 = frac{2sqrt{2}}{3}).3. Finding coordinates of point (C): - The distance (BC = 1) gives the equation: (sqrt{(x_2 + frac{7}{6})^2 + (y_2 - frac{2sqrt{2}}{3})^2} = 1). - Solving this with the semicircle equation (x_2^2 + y_2^2 = 2.25): - After solving, we find (x_2 = -frac{17}{54}) and (y_2 = frac{28sqrt{2}}{27}).4. Calculating the distance (CD): - Coordinates of (C) are (left(-frac{17}{54}, frac{28sqrt{2}}{27}right)) and (D) is ((1.5, 0)). - Using the distance formula: [ CD = sqrt{left(1.5 + frac{17}{54}right)^2 + left(0 - frac{28sqrt{2}}{27}right)^2} ] - Simplifying, we find: [ CD = sqrt{left(frac{49}{27}right)^2 + left(frac{28sqrt{2}}{27}right)^2} = sqrt{frac{3969}{729}} = frac{63}{27} = frac{7}{3} ]The length (CD) is (boxed{dfrac{7}{3}}).,<think>Okay, so I have this geometry problem here. Let me try to visualize it first. There's a semicircle with diameter AD, which is 3 units long. Points B and C are on the semicircle such that AB = BC = 1. I need to find the length of CD. Hmm, let me start by drawing a diagram to get a better idea.First, I'll draw a straight line AD of length 3. Since it's a semicircle, the center must be at the midpoint of AD. Let's calculate the midpoint. If AD is 3, then the midpoint, let's call it O, is at 1.5 units from both A and D. So, AO = OD = 1.5. The semicircle is then drawn with radius 1.5, right?Now, points B and C are on the semicircle. The distances from A to B and from B to C are both 1. So, starting from A, moving 1 unit along the semicircle to point B, then another 1 unit to point C, and then from C to D, which is the remaining distance. I need to find CD.Wait, but how exactly are these points placed? Since it's a semicircle, all points are on the circumference. So AB and BC are chords of the semicircle, each of length 1. Let me recall that in a circle, the length of a chord is related to the central angle subtended by the chord. The formula for the length of a chord is ( 2r sin frac{theta}{2} ), where ( theta ) is the central angle in radians and r is the radius.Given that the radius r here is 1.5, since AD is 3, so radius is 1.5. So, for chord AB of length 1, the central angle ( theta_1 ) satisfies ( 1 = 2 times 1.5 times sin frac{theta_1}{2} ). Let me write that down:( AB = 2r sin frac{theta_1}{2} )So,( 1 = 3 sin frac{theta_1}{2} )Therefore,( sin frac{theta_1}{2} = frac{1}{3} )So,( frac{theta_1}{2} = arcsin left( frac{1}{3} right) )Hence,( theta_1 = 2 arcsin left( frac{1}{3} right) )Similarly, for chord BC, which is also length 1, the central angle ( theta_2 ) would be the same as ( theta_1 ), right? Because BC is another chord of length 1 in the same circle, so the central angle should be the same.Therefore, each of AB and BC corresponds to a central angle of ( 2 arcsin left( frac{1}{3} right) ). Then, the total central angle from A to C would be ( theta_1 + theta_2 = 4 arcsin left( frac{1}{3} right) ). Then, the remaining central angle from C to D would be the total semicircle angle, which is ( pi ) radians, minus ( 4 arcsin left( frac{1}{3} right) ).Therefore, the central angle for CD is ( pi - 4 arcsin left( frac{1}{3} right) ). Then, the length of chord CD can be calculated using the same chord length formula:( CD = 2r sin frac{theta_3}{2} )Where ( theta_3 = pi - 4 arcsin left( frac{1}{3} right) ), so( CD = 3 sin left( frac{pi}{2} - 2 arcsin left( frac{1}{3} right) right) )Wait, because ( frac{theta_3}{2} = frac{pi}{2} - 2 arcsin left( frac{1}{3} right) )Using the sine of a difference identity:( sin left( frac{pi}{2} - 2 arcsin left( frac{1}{3} right) right) = cos left( 2 arcsin left( frac{1}{3} right) right) )So,( CD = 3 cos left( 2 arcsin left( frac{1}{3} right) right) )Now, let's compute ( cos left( 2 arcsin left( frac{1}{3} right) right) ). There's a double-angle formula for cosine in terms of sine:( cos(2theta) = 1 - 2 sin^2 theta )Let ( theta = arcsin left( frac{1}{3} right) ). Then,( cos left( 2 theta right) = 1 - 2 left( frac{1}{3} right)^2 = 1 - frac{2}{9} = frac{7}{9} )Therefore,( CD = 3 times frac{7}{9} = frac{21}{9} = frac{7}{3} )Wait, but hold on. If CD is 7/3, that's approximately 2.333, but the entire diameter is 3. So from A to D is 3, and if AB and BC are each 1, then CD should be 3 - 1 -1 = 1? But that's along the diameter. But the chords AB, BC, CD are along the semicircle, not the diameter. So the chord CD is not directly the remaining length on the diameter.So my previous calculation gives CD as 7/3, which is approximately 2.333, which seems plausible because it's a chord, not the straight line. Let me verify this.Alternatively, maybe I can use coordinate geometry. Let's place the semicircle on a coordinate system with the center O at (1.5, 0), diameter AD along the x-axis from (0,0) to (3,0). The semicircle is the upper half.Points A is at (0,0), D at (3,0). Points B and C are on the semicircle with coordinates (x1, y1) and (x2, y2), respectively. The distances AB and BC are 1 each. Let me parametrize points B and C.Since AB = 1, point B is 1 unit away from A(0,0). The coordinates of B must satisfy both the distance from A and lying on the semicircle.So, equations:For point B:1. ( (x_B - 0)^2 + (y_B - 0)^2 = 1^2 )2. ( (x_B - 1.5)^2 + y_B^2 = (1.5)^2 )Similarly, for point C:Distance from B to C is 1:3. ( (x_C - x_B)^2 + (y_C - y_B)^2 = 1^2 )And point C is on the semicircle:4. ( (x_C - 1.5)^2 + y_C^2 = (1.5)^2 )So, solving equations 1 and 2 for point B.From equation 1: ( x_B^2 + y_B^2 = 1 )From equation 2: ( (x_B - 1.5)^2 + y_B^2 = 2.25 )Subtract equation 1 from equation 2:( (x_B - 1.5)^2 + y_B^2 - x_B^2 - y_B^2 = 2.25 - 1 )Simplify:( x_B^2 - 3x_B + 2.25 - x_B^2 = 1.25 )So,( -3x_B + 2.25 = 1.25 )( -3x_B = -1 )( x_B = frac{1}{3} )Then, substitute x_B back into equation 1:( left( frac{1}{3} right)^2 + y_B^2 = 1 )( frac{1}{9} + y_B^2 = 1 )( y_B^2 = frac{8}{9} )Since it's the upper semicircle, y_B is positive:( y_B = frac{2sqrt{2}}{3} )So, point B is at ( left( frac{1}{3}, frac{2sqrt{2}}{3} right) )Now, let's find point C. It is 1 unit away from B and also on the semicircle. Let me denote point C as (x_C, y_C). Then:From equation 3:( left( x_C - frac{1}{3} right)^2 + left( y_C - frac{2sqrt{2}}{3} right)^2 = 1 )From equation 4:( left( x_C - frac{3}{2} right)^2 + y_C^2 = left( frac{3}{2} right)^2 = frac{9}{4} )Let me expand equation 4:( x_C^2 - 3x_C + frac{9}{4} + y_C^2 = frac{9}{4} )Simplify:( x_C^2 - 3x_C + y_C^2 = 0 )Equation 3 expanded:( x_C^2 - frac{2}{3}x_C + frac{1}{9} + y_C^2 - frac{4sqrt{2}}{3}y_C + frac{8}{9} = 1 )Combine terms:( x_C^2 + y_C^2 - frac{2}{3}x_C - frac{4sqrt{2}}{3}y_C + frac{1}{9} + frac{8}{9} - 1 = 0 )Simplify constants:( frac{1}{9} + frac{8}{9} = 1 ), so 1 -1 = 0. So equation 3 becomes:( x_C^2 + y_C^2 - frac{2}{3}x_C - frac{4sqrt{2}}{3}y_C = 0 )Now, from equation 4, we have ( x_C^2 + y_C^2 = 3x_C ). Let me substitute this into equation 3.So replace ( x_C^2 + y_C^2 ) with ( 3x_C ):( 3x_C - frac{2}{3}x_C - frac{4sqrt{2}}{3}y_C = 0 )Simplify:( left( 3 - frac{2}{3} right)x_C - frac{4sqrt{2}}{3}y_C = 0 )( frac{7}{3}x_C - frac{4sqrt{2}}{3}y_C = 0 )Multiply both sides by 3:( 7x_C - 4sqrt{2} y_C = 0 )So,( 7x_C = 4sqrt{2} y_C )Therefore,( y_C = frac{7}{4sqrt{2}} x_C = frac{7sqrt{2}}{8} x_C )Now, substitute this into equation 4:( x_C^2 + y_C^2 = 3x_C )Plugging in y_C:( x_C^2 + left( frac{7sqrt{2}}{8} x_C right)^2 = 3x_C )Calculate ( left( frac{7sqrt{2}}{8} x_C right)^2 ):( frac{49 times 2}{64} x_C^2 = frac{98}{64} x_C^2 = frac{49}{32} x_C^2 )So equation becomes:( x_C^2 + frac{49}{32} x_C^2 = 3x_C )Combine like terms:( left( 1 + frac{49}{32} right) x_C^2 = 3x_C )Convert 1 to 32/32:( frac{32 + 49}{32} x_C^2 = 3x_C )( frac{81}{32} x_C^2 = 3x_C )Multiply both sides by 32:( 81x_C^2 = 96x_C )Divide both sides by x_C (assuming x_C ≠ 0, which it can't be because point C is not at A or D):( 81x_C = 96 )Therefore,( x_C = frac{96}{81} = frac{32}{27} approx 1.185 )Then, y_C = (7√2 / 8) * x_C = (7√2 / 8) * (32/27) = (7√2 * 32) / (8 * 27) = (7√2 * 4) / 27 = (28√2)/27So, point C is at (32/27, 28√2/27)Now, need to find CD. Point D is at (3, 0). So, distance CD is:√[(3 - 32/27)^2 + (0 - 28√2/27)^2]Calculate 3 - 32/27 = (81/27 - 32/27) = 49/27So, the x-component squared: (49/27)^2 = 2401/729The y-component squared: (28√2/27)^2 = (784 * 2)/729 = 1568/729Total CD^2 = 2401/729 + 1568/729 = (2401 + 1568)/729 = 3969/729Simplify 3969 ÷ 729. Let's see: 729 * 5 = 3645, 3969 - 3645 = 324, which is 729 * 0.444… Wait, 729 * 5.5 = 729*5 + 729*0.5 = 3645 + 364.5 = 4009.5. Hmm, that's too much. Alternatively, 3969 ÷ 729. Since 729=9^3=27^2. 3969 ÷ 729. Let's divide numerator and denominator by 9: 441/81. Divide by 9 again: 49/9. So, 49/9. Therefore, CD^2 = 49/9, so CD = 7/3 ≈ 2.333Which matches the previous result. So CD is 7/3. That seems correct.But let me check once more. So starting from A, moving along the semicircle 1 unit to B, then another 1 unit to C, then CD is 7/3. Since the total circumference of the semicircle is πr = π*1.5 ≈ 4.712. The length from A to B to C to D would be 1 +1 +7/3 ≈ 1 +1 +2.333≈4.333, which is less than the semicircle's circumference. That seems plausible because the chord lengths are straight lines, not arc lengths. Wait, hold on, chord length vs arc length. Wait, in the problem statement, it says "two points B and C are placed such that AB=BC=1". So AB and BC are straight-line distances (chords), not arc lengths. So the chord lengths are 1 each, so the arc lengths would be longer. So the total chord lengths from A to B to C to D sum to 1 +1 +7/3 ≈ 4.333, which is less than the diameter 3, but chord lengths are not along the diameter. So the answer CD=7/3 seems correct.Alternatively, if I had considered arc lengths, the approach would have been different, but the problem specifies AB=BC=1, which are straight-line distances. Therefore, the chord length CD is indeed 7/3. So I think the answer is boxed{dfrac{7}{3}}.Final AnswerThe length ( CD ) is boxed{dfrac{7}{3}}.</think>Given a semicircle with diameter (AD = 3) and radius (1.5), points (B) and (C) are placed such that (AB = BC = 1). We need to find the length (CD).1. Central Angles Calculation: - The length of chord (AB) is 1. Using the chord length formula (2r sin frac{theta}{2}), we find the central angle (theta_1) for (AB): [ 1 = 2 times 1.5 times sin frac{theta_1}{2} implies sin frac{theta_1}{2} = frac{1}{3} implies theta_1 = 2 arcsin left( frac{1}{3} right) ] - Similarly, the central angle (theta_2) for (BC) is the same as (theta_1).2. Remaining Central Angle: - The total central angle from (A) to (C) is (2 theta_1 = 4 arcsin left( frac{1}{3} right)). - The remaining central angle from (C) to (D) is: [ pi - 4 arcsin left( frac{1}{3} right) ]3. Length of Chord (CD): - Using the chord length formula for (CD) with the central angle (pi - 4 arcsin left( frac{1}{3} right)): [ CD = 2 times 1.5 times sin left( frac{pi - 4 arcsin left( frac{1}{3} right)}{2} right) = 3 cos left( 2 arcsin left( frac{1}{3} right) right) ] - Using the double-angle identity for cosine: [ cos left( 2 arcsin left( frac{1}{3} right) right) = 1 - 2 left( frac{1}{3} right)^2 = frac{7}{9} ] - Therefore, the length (CD) is: [ CD = 3 times frac{7}{9} = frac{7}{3} ]4. Verification with Coordinate Geometry: - Placing the semicircle on a coordinate system, solving for coordinates of (B) and (C), and calculating the distance (CD) confirms the result.The length (CD) is (boxed{dfrac{7}{3}}).]

🔑:Let H be the intersection of [A C] with [O B], where O is the center of the circle. We have C D = sqrt{A D^{2} - A C^{2}} with A C = 2 A H.Let K be the midpoint of [A B]. The triangle A O B is isosceles at O. We denote alpha = widehat{H A B} = widehat{B O K}. We have cos alpha = A H / A B = O K / O B so A H = frac{2}{3} O K. Furthermore, O K = sqrt{O A^{2} - A K^{2}} = sqrt{(3 / 2)^{2} - (1 / 2)^{2}} = sqrt{2}, thus A C = frac{4}{3} sqrt{2}, and finally C D = sqrt{9 - 32 / 9} = 7 / 3.

🧠:[<think>Okay, so I have this problem here: Given the function y = lg(4 - x) with its domain as set A, and set B = {x | x < a}. The question is asking for the range of the real number a such that "x belongs to A" is a sufficient but not necessary condition for "x belongs to B". Hmm, let me try to parse this step by step.First, let's recall what the domain of a logarithmic function is. The function lg(4 - x) is defined when the argument inside the logarithm is positive. So, 4 - x > 0. Solving that inequality gives x < 4. Therefore, the domain A is all real numbers x such that x < 4. So, set A is (-∞, 4).Set B is given as {x | x < a}, which is just the interval (-∞, a). Now, the condition is that "x belongs to A" is a sufficient but not necessary condition for "x belongs to B". In logical terms, this means that if x is in A, then x must be in B (sufficient), but there are elements in B that are not in A (not necessary). Let me rephrase that. If A is a sufficient condition for B, then every element of A is also an element of B. So, A is a subset of B. However, since it's not necessary, B is not a subset of A, meaning there are elements in B that are not in A. Therefore, A must be a proper subset of B. Wait, no. Let me think again. If "x belongs to A" implies "x belongs to B", then A is a subset of B. But since it's not necessary, the converse isn't true. So, B is not a subset of A. Therefore, A is a subset of B, but B is not a subset of A. Which would mean that A is a proper subset of B. So, all elements of A are in B, but there are elements in B not in A. Therefore, B must be a superset of A. So, set A is (-∞, 4) and set B is (-∞, a). For A to be a subset of B, the upper bound of A (which is 4) must be less than or equal to the upper bound of B (which is a). Wait, but upper bounds here. If A is (-∞,4), and B is (-∞,a), then for A to be a subset of B, every x < 4 must also satisfy x < a. That would mean that 4 must be less than or equal to a. Because if a were less than 4, then there would be elements in A (like x = a) that are not in B. For example, if a were 3, then A is up to 4, but B is up to 3. So, numbers between 3 and 4 would be in A but not in B, making A not a subset of B. Therefore, to have A as a subset of B, a must be at least 4. But then, if a is equal to 4, then B would be (-∞,4), same as A. So, in that case, A is a subset of B, but B is equal to A. However, the problem states that "x belongs to A" is a sufficient but not necessary condition. If A and B are equal, then "x belongs to A" would be both necessary and sufficient for "x belongs to B". Therefore, to have it not necessary, B must contain elements not in A, so B must be a proper superset of A. Therefore, a must be greater than 4. Wait, but if a is greater than 4, then B is (-∞, a), which includes all of A (since A is up to 4) and more (from 4 to a). So, in this case, A is a subset of B, and B has elements not in A. Therefore, the sufficient condition (A) does not require all elements of B, hence the necessary part is not met. So, that fits the problem's requirement. Therefore, the range of a should be a > 4. But let me verify again.Original problem: "x belongs to A" is a sufficient but not necessary condition for "x belongs to B".In logic terms, this translates to: A implies B (sufficient), but B does not imply A (not necessary). So, A is a subset of B, but B is not a subset of A. Which is exactly the case when A is a proper subset of B. So, A must be entirely contained in B, but B is larger. Since A is (-∞,4), and B is (-∞,a), then for A to be a subset of B, a must be greater than 4. If a is 4, then B is equal to A, so A is a subset of B, but B is not a superset with more elements. So, in that case, being in A is equivalent to being in B, which would make the conditions necessary and sufficient. Therefore, to have it only sufficient but not necessary, a must be greater than 4. Therefore, the range of a is (4, ∞). So, the answer should be a > 4, which in interval notation is (4, ∞). Therefore, the range of the real number a is all real numbers greater than 4. Wait, but let me check once more. Suppose a = 5. Then B is (-∞,5). Then, if x is in A (x < 4), then x is automatically in B (x < 5). So, A is a subset of B. But there are elements in B not in A, like x = 4.5, which is in B but not in A. Therefore, the condition is sufficient but not necessary. If a = 4, then B is (-∞,4), same as A. So, x in A is equivalent to x in B. Therefore, "x in A" would be necessary and sufficient. So, that's not what the problem wants. If a < 4, say a = 3, then B is (-∞,3). Then, A is (-∞,4). So, elements in A like x = 3.5 are in A but not in B. Therefore, A is not a subset of B, so the implication fails. Therefore, a must be greater than 4. Therefore, the answer should be a > 4, so the range of a is (4, ∞). Therefore, the real number a must be greater than 4. I think that's it. Let me just recap. The key points are understanding the definitions of sufficient and necessary conditions. Since A needs to be a subset of B (sufficient), but B should not be a subset of A (not necessary). Therefore, A must be strictly contained in B, which translates to the upper limit of B (a) being greater than the upper limit of A (4). So, a > 4. Yep, that makes sense. Final AnswerThe range of the real number ( a ) is boxed{(4, +infty)}.</think>Given the function ( y = lg(4 - x) ), the domain ( A ) is determined by the condition that the argument of the logarithm must be positive. This gives ( 4 - x > 0 ), which simplifies to ( x < 4 ). Therefore, the domain ( A ) is the interval ( (-infty, 4) ).Set ( B ) is defined as ( { x mid x < a } ), which is the interval ( (-infty, a) ).The problem states that "x belongs to A" is a sufficient but not necessary condition for "x belongs to B". This means:1. If ( x in A ), then ( x in B ) (sufficient condition).2. There exist elements in ( B ) that are not in ( A ) (not necessary condition).For ( A ) to be a subset of ( B ), all elements of ( A ) must be in ( B ). This requires ( a ) to be greater than 4. If ( a leq 4 ), then ( B ) would not contain all elements of ( A ), failing the sufficient condition. If ( a = 4 ), ( B ) would be equal to ( A ), making the condition both necessary and sufficient, which is not required. Therefore, ( a ) must be greater than 4 to ensure ( B ) contains all elements of ( A ) and additional elements not in ( A ).Thus, the range of the real number ( a ) is (boxed{(4, +infty)}).,<think>Okay, let me try to work through this problem step by step. So, the problem is about finding the range of the real number ( a ) such that the condition "x belongs to A" is sufficient but not necessary for "x belongs to B". First, let me parse the problem again to make sure I understand everything correctly. We have the function ( y = lg(4 - x) ), and its domain is set ( A ). Then set ( B = {x | x < a} ). The key part here is understanding the logical condition: "x belongs to A" is a sufficient but not necessary condition for "x belongs to B". Alright, so in logical terms, if "x ∈ A" is a sufficient condition for "x ∈ B", that means whenever x is in A, it must also be in B. In other words, A is a subset of B. But it's also stated that it's not a necessary condition, which means that there are elements in B that are not in A. So, B has to be a superset of A but not the other way around. Therefore, A must be a proper subset of B. So, translating that into set theory terms, we need ( A subseteq B ) but ( B nsubseteq A ). Which would mean every element of A is in B, but there are elements in B that are not in A. Therefore, A is strictly contained within B. Now, let's figure out what set A is. The function given is ( y = lg(4 - x) ). Remember that the logarithm function ( lg ) is defined only when its argument is positive. So, the domain ( A ) of this function is all real numbers ( x ) such that ( 4 - x > 0 ). Solving that inequality:( 4 - x > 0 Rightarrow -x > -4 Rightarrow x < 4 ).So, set ( A ) is ( (-infty, 4) ).Set ( B ) is given as ( {x | x < a} ), which is the interval ( (-infty, a) ).Now, the condition is that ( A subseteq B ), so every element in A must be in B. Since A is ( (-infty, 4) ), for this interval to be a subset of ( (-infty, a) ), the upper bound of A (which is 4) must be less than or equal to the upper bound of B (which is a). Therefore, we have ( 4 leq a ).But wait, if ( a = 4 ), then B would be ( (-infty, 4) ), which is exactly the same as A. But the problem states that "x belongs to A" is a sufficient but not necessary condition for "x belongs to B". If A and B are equal, then "x belongs to A" would be both necessary and sufficient for "x belongs to B", right? So, in that case, the condition would not be only sufficient but not necessary. Therefore, ( a ) cannot be equal to 4. So, ( a ) must be greater than 4. Wait, but hold on. Let me verify this again. If ( a > 4 ), then B is ( (-infty, a) ), which includes all elements of A (since A is up to 4) and also elements from 4 to a. So in this case, A is a subset of B, and B has elements not in A. Therefore, "x ∈ A" implies "x ∈ B" (sufficient), but "x ∈ B" does not imply "x ∈ A" (not necessary). So that's exactly the condition we need.However, if ( a = 4 ), then A and B are the same set. Therefore, "x ∈ A" would be equivalent to "x ∈ B", making the conditions both necessary and sufficient. So, that's not allowed here because the problem specifies that it's only sufficient, not necessary. Therefore, ( a ) has to be greater than 4.But let's check the other side. What if ( a < 4 )? Then, B would be ( (-infty, a) ), but A is ( (-infty, 4) ). So, in this case, B is a subset of A. Therefore, "x ∈ A" does not imply "x ∈ B" because there are elements in A (specifically between a and 4) that are not in B. So, this violates the sufficiency condition. Therefore, ( a ) can't be less than 4.Therefore, the only possibility is ( a > 4 ). So, the range of the real number ( a ) is all real numbers greater than 4, which in interval notation is ( (4, infty) ).But let me make sure I didn't miss anything here. Let me think through it again.Set A is ( (-infty, 4) ), set B is ( (-infty, a) ). For A to be a subset of B, every element of A must be in B. The supremum of A is 4, so B must include all numbers less than 4. Therefore, the upper limit of B, which is a, must be at least 4. But if a is exactly 4, then B is ( (-infty, 4) ), same as A. So, in that case, A is a subset of B (they are equal), but then "x ∈ A" is equivalent to "x ∈ B", which would make them necessary and sufficient for each other. But the problem states that "x ∈ A" is sufficient but not necessary for "x ∈ B", which means that there must exist elements in B that are not in A. If a is 4, there are no such elements. Therefore, a must be greater than 4. Therefore, the range of a is ( a > 4 ).Yes, that seems correct. So, the answer is ( (4, +infty) ), which in boxed form is boxed{(4, +infty)}.Final AnswerThe range of the real number ( a ) is boxed{(4, +infty)}.</think>Given the function ( y = lg(4 - x) ), its domain ( A ) is determined by the requirement that the argument of the logarithm be positive. Solving ( 4 - x > 0 ), we find ( x < 4 ). Therefore, the domain ( A ) is ( (-infty, 4) ).Set ( B ) is defined as ( {x mid x < a} ), which is the interval ( (-infty, a) ).For "x belongs to A" to be a sufficient but not necessary condition for "x belongs to B", ( A ) must be a proper subset of ( B ). This means every element of ( A ) must be in ( B ), but there must be elements in ( B ) that are not in ( A ).To satisfy ( A subseteq B ), the upper bound of ( A ) (which is 4) must be less than or equal to the upper bound of ( B ) (which is ( a )). However, if ( a = 4 ), then ( A ) and ( B ) would be equal, making "x belongs to A" both necessary and sufficient, which contradicts the requirement. Therefore, ( a ) must be greater than 4.Thus, the range of the real number ( a ) is (boxed{(4, +infty)}).]

🔑:For the function y = lg(4-x), its domain A is determined by the condition 4-x > 0, which simplifies to x < 4. Therefore, for "x belongs to A" to be a sufficient but not necessary condition for "x belongs to B", it means that every element in A must also be in B, but not every element in B has to be in A. This implies that a must be greater than 4. Hence, the range of the real number a is boxed{(4, +infty)}.

🧠:[<think>Okay, let's try to tackle this problem step by step. There are three parts here, so I need to go through each one carefully. Let me start with part 1.Problem 1: Find the equation of ellipse C.The given equation of the ellipse is (dfrac{x^2}{2t^2} + dfrac{y^2}{t^2} = 1) with (t > 0). The foci (F_1) and (F_2) are the left and right foci, respectively. The minimum distance from a point on the ellipse to (F_2) is (2sqrt{2} - 2). We need to find the equation of the ellipse, which probably means finding the value of (t) first.First, recall that for an ellipse (dfrac{x^2}{a^2} + dfrac{y^2}{b^2} = 1), the distance between the center and each focus is (c = sqrt{a^2 - b^2}) when (a > b). But in our case, the denominators are (2t^2) and (t^2). Let me check which is larger. Since (2t^2 > t^2) for (t > 0), so the major axis is along the x-axis. Therefore, (a^2 = 2t^2) and (b^2 = t^2), so (a = tsqrt{2}) and (b = t). Then, the distance from the center to each focus is (c = sqrt{a^2 - b^2} = sqrt{2t^2 - t^2} = sqrt{t^2} = t). Therefore, the foci are at ((-c, 0)) and ((c, 0)), which would be ((-t, 0)) and ((t, 0)), so (F_1 = (-t, 0)) and (F_2 = (t, 0)).Now, the minimum distance from a point on the ellipse to (F_2) is given as (2sqrt{2} - 2). Let's recall that the minimum distance from a focus to the ellipse is (a - c). Wait, is that correct? For an ellipse, the closest point to a focus is along the major axis towards that focus. So, since the major axis is along the x-axis, the closest point to (F_2) would be the rightmost vertex of the ellipse, which is at ((a, 0)). The distance from (F_2) (which is at ((t, 0))) to this vertex ((a, 0)) is (a - t). Wait, but (a = tsqrt{2}), so the distance is (tsqrt{2} - t = t(sqrt{2} - 1)). According to the problem, this minimum distance is (2sqrt{2} - 2). Therefore:(t(sqrt{2} - 1) = 2sqrt{2} - 2)Let me solve for (t). Let's factor out (sqrt{2} - 1) on the right side:(2sqrt{2} - 2 = 2(sqrt{2} - 1))Therefore, (t(sqrt{2} - 1) = 2(sqrt{2} - 1))Divide both sides by (sqrt{2} - 1) (since it's positive, we can do that):(t = 2)So, (t = 2). Therefore, substituting back into the ellipse equation:(dfrac{x^2}{2*(2)^2} + dfrac{y^2}{(2)^2} = 1)Simplify:(dfrac{x^2}{8} + dfrac{y^2}{4} = 1)Therefore, the equation of ellipse C is (dfrac{x^2}{8} + dfrac{y^2}{4} = 1). Let me check if this makes sense. The major axis is along x with (a = 2sqrt{2}), so the vertices at ((pm 2sqrt{2}, 0)), and foci at ((pm t, 0) = (pm 2, 0)). Then the distance from (F_2 = (2, 0)) to the right vertex ((2sqrt{2}, 0)) is (2sqrt{2} - 2), which is indeed the given minimum distance. That checks out. So part 1 is solved.Problem 2: When the dot product (vec{F_1N} cdot vec{F_2N} = 0), find the area of triangle (F_1MN).Okay, so first, we need to parse the given condition. Points M and N are on the ellipse above the x-axis. Vector (vec{F_1M}) is parallel to (vec{F_2N}). Additionally, the dot product of (vec{F_1N}) and (vec{F_2N}) is zero, which means they are perpendicular. From this, we need to find the area of triangle (F_1MN).First, let's recap the ellipse equation we found: (dfrac{x^2}{8} + dfrac{y^2}{4} = 1). Therefore, (a = 2sqrt{2}), (b = 2), and foci at ((pm 2, 0)). So, (F_1 = (-2, 0)), (F_2 = (2, 0)).Points M and N are on the ellipse above the x-axis, so their coordinates are ((x_m, y_m)) and ((x_n, y_n)) with (y_m, y_n > 0). The vectors (vec{F_1M}) and (vec{F_2N}) are parallel. Let's express these vectors.(vec{F_1M} = (x_m - (-2), y_m - 0) = (x_m + 2, y_m))(vec{F_2N} = (x_n - 2, y_n - 0) = (x_n - 2, y_n))Parallel vectors mean one is a scalar multiple of the other. So, there exists a scalar (k) such that:(x_m + 2 = k(x_n - 2))(y_m = k y_n)Additionally, since M and N are on the ellipse, they satisfy:(dfrac{x_m^2}{8} + dfrac{y_m^2}{4} = 1)(dfrac{x_n^2}{8} + dfrac{y_n^2}{4} = 1)Also, the condition (vec{F_1N} cdot vec{F_2N} = 0). Let's compute those vectors.(vec{F_1N} = (x_n - (-2), y_n - 0) = (x_n + 2, y_n))(vec{F_2N} = (x_n - 2, y_n))Their dot product is:((x_n + 2)(x_n - 2) + y_n * y_n = 0)Simplify:((x_n^2 - 4) + y_n^2 = 0)So,(x_n^2 + y_n^2 - 4 = 0)But we also know that (dfrac{x_n^2}{8} + dfrac{y_n^2}{4} = 1). Let me write that as:Multiply both sides by 8: (x_n^2 + 2y_n^2 = 8)So we have two equations:1. (x_n^2 + y_n^2 = 4) (from the dot product condition)2. (x_n^2 + 2y_n^2 = 8) (from the ellipse equation)Subtract equation 1 from equation 2:( (x_n^2 + 2y_n^2) - (x_n^2 + y_n^2) = 8 - 4 )( y_n^2 = 4 )Since point N is above the x-axis, (y_n > 0), so (y_n = 2).Substitute back into equation 1: (x_n^2 + (2)^2 = 4) => (x_n^2 = 0) => (x_n = 0). Wait, but if (x_n = 0), then point N is (0, 2). Let's check if this is on the ellipse: (0^2/8 + 2^2/4 = 0 + 1 = 1. Yes, that's valid. So N is (0, 2).Now, we need to find point M such that (vec{F_1M}) is parallel to (vec{F_2N}). Let's compute (vec{F_2N}):Since N is (0, 2), then (vec{F_2N} = (0 - 2, 2 - 0) = (-2, 2)). Therefore, (vec{F_1M}) must be a scalar multiple of (-2, 2). Let me denote the scalar as k. So:(vec{F_1M} = k vec{F_2N} = k(-2, 2) = (-2k, 2k))But (vec{F_1M}) is also equal to (x_m + 2, y_m). So:x_m + 2 = -2ky_m = 2kTherefore, x_m = -2k - 2y_m = 2kSince point M is on the ellipse:(dfrac{(-2k - 2)^2}{8} + dfrac{(2k)^2}{4} = 1)Let's expand this:First term: [ (4k^2 + 8k + 4) ] / 8Second term: [4k^2] / 4 = k^2So:(4k^2 + 8k + 4)/8 + k^2 = 1Simplify fractions:First term: (4k^2)/8 + 8k/8 + 4/8 = 0.5k^2 + k + 0.5Second term: k^2Adding them: 0.5k^2 + k + 0.5 + k^2 = 1.5k^2 + k + 0.5 = 1So:1.5k^2 + k + 0.5 = 1Subtract 1:1.5k^2 + k - 0.5 = 0Multiply both sides by 2 to eliminate decimals:3k^2 + 2k - 1 = 0Now, solve quadratic equation:k = [-2 ± sqrt(4 + 12)] / (2*3) = [-2 ± sqrt(16)] / 6 = [-2 ± 4]/6So two solutions:k = (-2 + 4)/6 = 2/6 = 1/3k = (-2 - 4)/6 = -6/6 = -1Now, check which k is valid.Since point M is above the x-axis, y_m = 2k > 0. So:For k = 1/3: y_m = 2*(1/3) = 2/3 > 0, valid.For k = -1: y_m = 2*(-1) = -2 < 0, which is below the x-axis. Since M must be above, we discard k = -1.Therefore, k = 1/3.Therefore, coordinates of M:x_m = -2*(1/3) - 2 = -2/3 - 2 = -8/3y_m = 2*(1/3) = 2/3Wait, but point M is (-8/3, 2/3). Let's check if this is on the ellipse:x_m^2 /8 + y_m^2 /4 = (64/9)/8 + (4/9)/4 = (8/9) + (1/9) = 9/9 = 1. Correct.So M is (-8/3, 2/3), N is (0, 2).Now, we need to find the area of triangle (F_1MN). The coordinates are:F1: (-2, 0)M: (-8/3, 2/3)N: (0, 2)Area can be calculated using the shoelace formula.Formula:Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Assigning:F1: (-2, 0) = (x1, y1)M: (-8/3, 2/3) = (x2, y2)N: (0, 2) = (x3, y3)Compute:1/2 | (-2)(2/3 - 2) + (-8/3)(2 - 0) + 0(0 - 2/3) |First compute each term:First term: (-2)(2/3 - 2) = (-2)(-4/3) = 8/3Second term: (-8/3)(2 - 0) = (-8/3)(2) = -16/3Third term: 0*(anything) = 0Sum: 8/3 -16/3 + 0 = -8/3Take absolute value and multiply by 1/2:1/2 * | -8/3 | = 1/2 * 8/3 = 4/3Therefore, the area is 4/3.Wait, but let me check the order of points. The shoelace formula depends on the order. If we list the points in the order F1, M, N, then the area is as calculated. Alternatively, if the points are ordered differently, the result should be the same. Let me verify:Alternatively, compute vectors F1M and F1N, then take the cross product.Vectors:F1M = M - F1 = (-8/3 + 2, 2/3 - 0) = (-2/3, 2/3)F1N = N - F1 = (0 + 2, 2 - 0) = (2, 2)Area is 1/2 | cross product of F1M and F1N |Cross product in 2D is scalar magnitude |(dx1 * dy2 - dx2 * dy1)|Compute:(-2/3)(2) - (2/3)(2) = (-4/3 - 4/3) = -8/3Absolute value is 8/3, times 1/2 is 4/3. Same result. So yes, area is 4/3.Thus, the answer to part 2 is 4/3.Problem 3: When (|vec{F_2N}| - |vec{F_1M}| = sqrt{6}), find the equation of the line (F_2N).Alright, so now we have another condition. Points M and N are still on the ellipse above the x-axis with (vec{F_1M}) parallel to (vec{F_2N}), but now the difference in their magnitudes is (sqrt{6}). We need to find the equation of line (F_2N).First, let's recall the setup. Points M and N are on the ellipse, above x-axis. (vec{F_1M} parallel vec{F_2N}), which as before, implies that (vec{F_1M} = k vec{F_2N}) for some scalar k. We need to use the condition (|vec{F_2N}| - |vec{F_1M}| = sqrt{6}). Then, once we find N, we can write the equation of line (F_2N).Let me denote again:(vec{F_1M} = (x_m + 2, y_m))(vec{F_2N} = (x_n - 2, y_n))From the parallelism, (vec{F_1M} = k vec{F_2N}), so:(x_m + 2 = k(x_n - 2))(y_m = k y_n)Also, points M and N are on the ellipse:(dfrac{x_m^2}{8} + dfrac{y_m^2}{4} = 1)(dfrac{x_n^2}{8} + dfrac{y_n^2}{4} = 1)Additionally, the condition (|vec{F_2N}| - |vec{F_1M}| = sqrt{6}). Let's compute these magnitudes.(|vec{F_2N}| = sqrt{(x_n - 2)^2 + y_n^2})(|vec{F_1M}| = sqrt{(x_m + 2)^2 + y_m^2})But since (vec{F_1M} = k vec{F_2N}), then:(|vec{F_1M}| = |k| |vec{F_2N}|). Let's denote ( |vec{F_2N}| = d ), then (|vec{F_1M}| = |k| d). The given condition is:(d - |k| d = sqrt{6})Factor d:(d (1 - |k|) = sqrt{6})So we need to express d and k in terms of variables.But perhaps it's better to parameterize the points. Let me attempt to set up equations.From the parallelism:(x_m + 2 = k(x_n - 2))(y_m = k y_n)Express (x_m) and (y_m) in terms of (x_n, y_n, k):(x_m = k(x_n - 2) - 2)(y_m = k y_n)Since M is on the ellipse:(dfrac{[k(x_n - 2) - 2]^2}{8} + dfrac{(k y_n)^2}{4} = 1)Similarly, N is on the ellipse:(dfrac{x_n^2}{8} + dfrac{y_n^2}{4} = 1)This seems complicated, but maybe we can parameterize N. Since N is on the ellipse above the x-axis, we can write N in terms of a parameter. Let me use parametric coordinates.For the ellipse (dfrac{x^2}{8} + dfrac{y^2}{4} = 1), parametric equations are:(x = 2sqrt{2} cos theta)(y = 2 sin theta)for some angle (theta). So N can be represented as ((2sqrt{2} cos theta, 2 sin theta)) with (0 < theta < pi) (since above x-axis).Then, point M is determined by the parallelism condition. Let's express (vec{F_1M} = k vec{F_2N}).First, compute (vec{F_2N}):(vec{F_2N} = (2sqrt{2} cos theta - 2, 2 sin theta - 0) = (2sqrt{2} cos theta - 2, 2 sin theta))Then, (vec{F_1M} = k vec{F_2N}), so:(x_m + 2 = k(2sqrt{2} cos theta - 2))(y_m = k(2 sin theta))Therefore, coordinates of M:(x_m = k(2sqrt{2} cos theta - 2) - 2)(y_m = 2k sin theta)But M is on the ellipse, so:(dfrac{[k(2sqrt{2} cos theta - 2) - 2]^2}{8} + dfrac{(2k sin theta)^2}{4} = 1)This seems very involved. Maybe there's a better way. Alternatively, perhaps express k in terms of theta?Alternatively, note that since (vec{F_1M}) is parallel to (vec{F_2N}), the direction ratios are proportional. So:(dfrac{x_m + 2}{x_n - 2} = dfrac{y_m}{y_n} = k)So, (y_m = k y_n) and (x_m = k(x_n - 2) - 2)Given that M and N are on the ellipse, we can set up equations. Let me let (y_n = 2 sin theta), as in the parameterization. Then, (x_n = 2sqrt{2} cos theta). So N is parameterized by theta.Then, from the above, (y_m = k * 2 sin theta)And (x_m = k (2sqrt{2} cos theta - 2) - 2)Since M is on the ellipse:(dfrac{[k(2sqrt{2} cos theta - 2) - 2]^2}{8} + dfrac{(2k sin theta)^2}{4} = 1)Let's expand this equation.First term:Let me denote A = k(2√2 cosθ - 2) - 2So,A = 2√2 k cosθ - 2k - 2Thus, A² = [2√2 k cosθ - 2k - 2]^2Second term:(2k sinθ)^2 /4 = 4k² sin²θ /4 = k² sin²θSo the equation becomes:[ (2√2 k cosθ - 2k - 2)^2 ] /8 + k² sin²θ = 1This seems complex. Maybe instead of parameterizing, try another approach.Alternatively, use the fact that the vectors are parallel to set up a relation between coordinates.Given (vec{F_1M} parallel vec{F_2N}), so:(dfrac{x_m + 2}{x_n - 2} = dfrac{y_m}{y_n})Which is equivalent to:((x_m + 2) y_n = (x_n - 2) y_m)But also, both M and N are on the ellipse. So:For M: (x_m^2 /8 + y_m^2 /4 = 1)For N: (x_n^2 /8 + y_n^2 /4 = 1)So we have three equations:1. ((x_m + 2) y_n = (x_n - 2) y_m)2. (x_m^2 /8 + y_m^2 /4 = 1)3. (x_n^2 /8 + y_n^2 /4 = 1)Additionally, the condition (|vec{F_2N}| - |vec{F_1M}| = sqrt{6})Let me compute (|vec{F_2N}| = sqrt{(x_n - 2)^2 + y_n^2})and (|vec{F_1M}| = sqrt{(x_m + 2)^2 + y_m^2})So the condition is:(sqrt{(x_n - 2)^2 + y_n^2} - sqrt{(x_m + 2)^2 + y_m^2} = sqrt{6})This is a system of four equations with four variables: x_m, y_m, x_n, y_n. Solving this system seems challenging, but maybe we can find a relationship between variables.Alternatively, use the relation from the parallel vectors. From equation 1:((x_m + 2) y_n = (x_n - 2) y_m)Let me denote this as equation 1.Also, from the ellipse equations, we can express y_m and y_n in terms of x_m and x_n:For point M: (y_m^2 = 4(1 - x_m^2 /8) = 4 - x_m^2 /2)Similarly, for point N: (y_n^2 = 4 - x_n^2 /2)But perhaps substituting these into equation 1:((x_m + 2) sqrt{4 - x_n^2 /2} = (x_n - 2) sqrt{4 - x_m^2 /2})This is getting complicated. Maybe square both sides to eliminate the square roots?Let me try:((x_m + 2)^2 (4 - x_n^2 /2) = (x_n - 2)^2 (4 - x_m^2 /2))But this leads to a complicated equation. Let me see if there's another approach.Alternatively, since (vec{F_1M}) is parallel to (vec{F_2N}), and given the symmetry of the ellipse, perhaps there's a relation that can be exploited. Let me consider specific coordinates or possible symmetries.Alternatively, note that if we consider the ellipse's properties, the sum of distances from any point on the ellipse to the foci is constant. For our ellipse, (a = 2sqrt{2}), so the sum is (2a = 4sqrt{2}).Therefore, for any point P on the ellipse, ( |PF_1| + |PF_2| = 4sqrt{2} ).But in this problem, we have points M and N, each on the ellipse. However, the condition involves ( |F_2N| - |F_1M| = sqrt{6} ). Let me see if we can relate |F_2N| and |F_1M|.But since both M and N are on the ellipse, for point N: ( |F_1N| + |F_2N| = 4sqrt{2} )Similarly, for point M: ( |F_1M| + |F_2M| = 4sqrt{2} )But how does this relate to the given condition?We have ( |F_2N| - |F_1M| = sqrt{6} ). Let me denote ( |F_2N| = d ), then ( |F_1M| = d - sqrt{6} ).But from the property of the ellipse, for point M: ( |F_1M| + |F_2M| = 4sqrt{2} ). So ( (d - sqrt{6}) + |F_2M| = 4sqrt{2} ). Therefore, ( |F_2M| = 4sqrt{2} - d + sqrt{6} ).Similarly, for point N: ( |F_1N| + d = 4sqrt{2} ), so ( |F_1N| = 4sqrt{2} - d ).But how does this help? Maybe not directly. Let's think differently.Since (vec{F_1M} parallel vec{F_2N}), their direction ratios are the same. Therefore, the slopes of these vectors should be equal.Compute the slope of (vec{F_1M}): slope = ( (y_m - 0)/(x_m - (-2)) = y_m / (x_m + 2) )Slope of (vec{F_2N}): slope = ( (y_n - 0)/(x_n - 2) = y_n / (x_n - 2) )Setting them equal:( y_m / (x_m + 2) = y_n / (x_n - 2) )Which is the same as equation 1.Perhaps express ( y_m = y_n (x_m + 2)/(x_n - 2) )Plugging this into M's ellipse equation:( x_m^2 /8 + [y_n (x_m + 2)/(x_n - 2)]^2 /4 = 1 )But this seems messy.Alternatively, use parametric coordinates for N and express M in terms of N.Let me proceed with the parameterization approach.Let me set N as ((2sqrt{2} cos theta, 2 sin theta)), as before. Then, compute (vec{F_2N}):(vec{F_2N} = (2sqrt{2} cos theta - 2, 2 sin theta))Then, (vec{F_1M}) is parallel to this vector. Therefore, point M can be expressed as:( M = F_1 + k vec{F_2N} = (-2, 0) + k(2sqrt{2} cos theta - 2, 2 sin theta) )Thus, coordinates of M:( x_m = -2 + k(2sqrt{2} cos theta - 2) )( y_m = 0 + k(2 sin theta) )Since M is on the ellipse, substitute into ellipse equation:(dfrac{[-2 + k(2sqrt{2} cos theta - 2)]^2}{8} + dfrac{[2k sin theta]^2}{4} = 1)Let me expand this:First term:Numerator: [ -2 + 2sqrt{2} k cos theta - 2k ]^2Let me write this as [ 2sqrt{2} k cos theta - 2k - 2 ]^2Expanding this:= (2sqrt{2} k cos theta)^2 + (-2k - 2)^2 + 2*(2sqrt{2}k cos theta)*(-2k - 2)= 8k² cos²θ + (4k² + 8k + 4) + 2*( -4sqrt{2}k² cosθ - 4sqrt{2}k cosθ )Simplify:= 8k² cos²θ + 4k² + 8k + 4 - 8sqrt{2}k² cosθ - 8sqrt{2}k cosθDivide by 8:= [8k² cos²θ + 4k² + 8k + 4 - 8√2 k² cosθ - 8√2 k cosθ]/8= k² cos²θ + 0.5k² + k + 0.5 - √2 k² cosθ - √2 k cosθSecond term:[4k² sin²θ]/4 = k² sin²θSo total equation:k² cos²θ + 0.5k² + k + 0.5 - √2 k² cosθ - √2 k cosθ + k² sin²θ = 1Combine like terms:k² (cos²θ + sin²θ) + 0.5k² - √2 k² cosθ + k + 0.5 - √2 k cosθ = 1Since cos²θ + sin²θ = 1:k² + 0.5k² - √2 k² cosθ + k + 0.5 - √2 k cosθ = 1Simplify:1.5k² - √2 k² cosθ + k + 0.5 - √2 k cosθ = 1Let me write this as:1.5k² - √2 k² cosθ - √2 k cosθ + k + 0.5 - 1 = 0Simplify constants:0.5 - 1 = -0.5Thus:1.5k² - √2 k² cosθ - √2 k cosθ + k - 0.5 = 0This equation seems complicated. Maybe factor out terms:Factor out k from some terms:k²(1.5 - √2 cosθ) + k(1 - √2 cosθ) - 0.5 = 0This is a quadratic in k:[1.5 - √2 cosθ]k² + [1 - √2 cosθ]k - 0.5 = 0Multiply all terms by 2 to eliminate decimals:[3 - 2√2 cosθ]k² + [2 - 2√2 cosθ]k - 1 = 0Let me denote ( c = cosθ ), so:(3 - 2√2 c)k² + (2 - 2√2 c)k -1 = 0This quadratic equation in k can be solved for k in terms of c. However, it's still complex.But we also have the condition that ( |vec{F_2N}| - |vec{F_1M}| = sqrt{6} ).Compute ( |vec{F_2N}| = sqrt{(2sqrt{2} cosθ - 2)^2 + (2 sinθ)^2} )Simplify:= sqrt[ (8 cos²θ - 8√2 cosθ + 4) + 4 sin²θ ]= sqrt[8 cos²θ - 8√2 cosθ + 4 + 4 sin²θ]= sqrt[8 cos²θ + 4 sin²θ - 8√2 cosθ + 4]Factor out 4 from the first two terms:= sqrt[4(2 cos²θ + sin²θ) -8√2 cosθ + 4]But 2 cos²θ + sin²θ = cos²θ +1. So:= sqrt[4(cos²θ +1) -8√2 cosθ +4]= sqrt[4 cos²θ +4 +4 -8√2 cosθ]= sqrt[4 cos²θ -8√2 cosθ +8]Factor 4:= sqrt[4(cos²θ - 2√2 cosθ + 2)]= 2 sqrt[cos²θ - 2√2 cosθ + 2]Similarly, compute |vec{F_1M}| = sqrt[(x_m + 2)^2 + y_m^2]But from earlier, coordinates of M:x_m = -2 + k(2√2 cosθ -2)Thus, x_m + 2 = k(2√2 cosθ -2)y_m = 2k sinθTherefore:|vec{F_1M}| = sqrt[ k²(2√2 cosθ -2)^2 + (2k sinθ)^2 ]= sqrt[ k²(8 cos²θ -8√2 cosθ +4) +4k² sin²θ ]= sqrt[ k²(8 cos²θ -8√2 cosθ +4 +4 sin²θ) ]= sqrt[ k²(8 cos²θ +4 sin²θ -8√2 cosθ +4) ]Similar to the previous computation for |vec{F_2N}|:We can factor:= sqrt[ k²*4(2 cos²θ + sin²θ -2√2 cosθ +1) ]Again, 2 cos²θ + sin²θ = cos²θ +1, so:= sqrt[ k²*4(cos²θ +1 -2√2 cosθ +1 ) ]= sqrt[ k²*4(cos²θ -2√2 cosθ +2 ) ]= 2k sqrt[cos²θ -2√2 cosθ +2 ]Therefore, the condition ( |vec{F_2N}| - |vec{F_1M}| = sqrt{6} ) becomes:2 sqrt[cos²θ -2√2 cosθ +2 ] - 2k sqrt[cos²θ -2√2 cosθ +2 ] = sqrt{6}Factor out 2 sqrt[cos²θ -2√2 cosθ +2 ]:2(1 - k) sqrt[cos²θ -2√2 cosθ +2 ] = sqrt{6}Divide both sides by 2:(1 - k) sqrt[cos²θ -2√2 cosθ +2 ] = sqrt{6}/2Square both sides to eliminate sqrt:(1 -k)^2 [cos²θ -2√2 cosθ +2 ] = 6/4 = 3/2So:(1 - 2k + k²)(cos²θ -2√2 cosθ +2 ) = 3/2This is getting very complicated. But remember that we also have the quadratic equation in k derived earlier:(3 - 2√2 c)k² + (2 - 2√2 c)k -1 = 0, where c = cosθPerhaps substitute c into this equation and relate it to the squared condition.Alternatively, perhaps set u = cosθ. Let me denote u = cosθ. Then sinθ = sqrt(1 - u²). Since θ is between 0 and pi, sinθ is non-negative.Then, the quadratic equation in k becomes:(3 - 2√2 u)k² + (2 - 2√2 u)k -1 = 0And the squared condition:(1 -k)^2 [u² -2√2 u +2 ] = 3/2Let me see if we can express k from the quadratic equation and substitute into the squared condition.Let me solve the quadratic equation for k:(3 - 2√2 u)k² + (2 - 2√2 u)k -1 = 0Let me denote A = 3 - 2√2 uB = 2 - 2√2 uC = -1Then:k = [-B ± sqrt(B² -4AC)] / (2A)Compute discriminant D:D = B² -4AC = (2 - 2√2 u)^2 -4*(3 - 2√2 u)*(-1)= 4 - 8√2 u + 8u² +4*(3 - 2√2 u)= 4 -8√2 u +8u² +12 -8√2 u= 16 -16√2 u +8u²Factor 8:= 8(2 - 2√2 u + u²)But 2 - 2√2 u + u² is the same as [u² - 2√2 u +2 ]Wait, exactly the same as the term in the squared condition!So D = 8(u² - 2√2 u + 2 )Then, k = [ -B ± sqrt(D) ] / (2A )= [ -(2 - 2√2 u) ± sqrt(8(u² - 2√2 u +2 )) ] / [2*(3 - 2√2 u) ]Simplify sqrt(8(...)):sqrt(8(...)) = 2*sqrt(2(...)).But let's keep it as is for now.Let me write this:k = [ -2 + 2√2 u ± sqrt(8(u² -2√2 u +2 )) ] / [ 2*(3 - 2√2 u) ]This is complicated, but perhaps substitute into the squared condition.Recall the squared condition:(1 -k)^2 [u² -2√2 u +2 ] = 3/2Let me express (1 -k) from this:(1 -k) = sqrt(3/(2 [u² -2√2 u +2 ]))But this is after squaring, which introduces ambiguity. However, since all terms are positive (since points are above x-axis, and k must be positive for y_m to be positive), we can take the positive square root.But perhaps instead, use the expression for k in terms of u.Alternatively, let me denote S = u² -2√2 u +2. Then, the squared condition becomes:(1 -k)^2 S = 3/2From the quadratic equation solution, k is expressed in terms of u and S.Given the complexity, perhaps we need to make an intelligent guess or find specific values of θ that satisfy the conditions.Alternatively, suppose that the line F2N has a certain slope, and we can find its equation once we find coordinates of N.But perhaps there's a smarter approach. Let me think.Recall that in problem 2, when the dot product was zero, we found specific points. Maybe in problem 3, there's also a specific configuration. Let me try specific positions.For instance, suppose that point N is at (0, 2) as in problem 2. Then, |F2N| would be the distance from (2,0) to (0,2), which is sqrt(4 +4) = sqrt(8) = 2√2. Then, point M would be as found earlier: (-8/3, 2/3). Then, |F1M| is the distance from (-2,0) to (-8/3, 2/3), which is sqrt( (-8/3 +6/3)^2 + (2/3)^2 ) = sqrt( (-2/3)^2 + (2/3)^2 ) = sqrt(4/9 +4/9) = sqrt(8/9) = 2√2 /3 ≈ 0.94. Then, |F2N| - |F1M| = 2√2 - 2√2/3 = (6√2 - 2√2)/3 = 4√2/3 ≈ 1.88, which is not sqrt(6) ≈ 2.45. So that doesn't fit.Alternatively, suppose point N is at (2√2, 0), the rightmost vertex. But N must be above x-axis, so y_n >0. So (2√2,0) is on the x-axis, invalid. Similarly, leftmost vertex (-2√2,0) is below x-axis.How about if N is at the top of the ellipse, (0,2). Wait, we did that. Another point: suppose theta is 45 degrees.Let me try theta = 45 degrees. Then cosθ = √2/2, sinθ = √2/2.Then N is (2√2*(√2/2), 2*(√2/2)) = (2*2/2, √2) = (2, √2)Compute |F2N|: distance from (2,0) to (2, √2) is sqrt(0 + (√2)^2) = sqrt(2) ≈1.414But wait, N is (2, √2). Check if this is on the ellipse: x²/8 + y²/4 = 4/8 + 2/4 = 0.5 +0.5=1. Yes.Compute vec(F2N): (2 -2, √2 -0) = (0, √2)Then vec(F1M) is parallel to this, so vec(F1M) = (0, √2)*k. Therefore, M = F1 + (0, √2 k) = (-2, 0) + (0, √2 k) = (-2, √2 k). Since M is on the ellipse:x_m² /8 + y_m² /4 = 4/8 + (2 k²)/4 = 0.5 + 0.5k² =1 => 0.5k² =0.5 =>k²=1 =>k=1 (since y_m >0, k=1). So M is (-2, √2). Check if this is above x-axis: yes. Then, |F1M| is the distance from (-2,0) to (-2, √2) which is sqrt(0 + (√2)^2) = sqrt(2). |F2N| is sqrt(2). Then |F2N| - |F1M| = sqrt(2) - sqrt(2) =0≠ sqrt(6). Doesn't work.Another try: suppose theta is such that N is (sqrt(8 - 2y_n^2), y_n). Maybe pick y_n = sqrt(2), so x_n^2 /8 + (2)/4 =1 =>x_n^2 /8 = 0.5 =>x_n^2=4 =>x_n=±2. Since above x-axis, x_n=2 or -2. But x_n=2, y_n= sqrt(2). Then N is (2, sqrt(2)), same as before. Similarly, x_n=-2, y_n= sqrt(2), but N is above x-axis. So x_n=-2, y_n=sqrt(2). Check if that's on the ellipse: 4/8 +2/4=0.5+0.5=1. Yes.Compute vec(F2N): (-2 -2, sqrt(2)-0)= (-4, sqrt(2))Then vec(F1M) is parallel to this. So vec(F1M) = k*(-4, sqrt(2))Coordinates of M: F1 + vec(F1M) = (-2,0) + (-4k, sqrt(2)k)= (-2 -4k, sqrt(2)k)Since M is on the ellipse:[(-2 -4k)^2]/8 + [2k²]/4 =1Expand:(4 + 16k + 16k²)/8 + (2k²)/4 =1Simplify:0.5 + 2k + 2k² +0.5k² =1Total: 0.5 +2k +2.5k² =1So 2.5k² +2k -0.5=0Multiply by 2: 5k² +4k -1=0Solutions:k = [-4 ± sqrt(16 +20)]/10 = [-4 ± sqrt(36)]/10 = [-4 ±6]/10Thus k= (2)/10=1/5 or k=(-10)/10=-1k=1/5: y_m = sqrt(2)*(1/5) >0. Valid. k=-1: y_m negative, invalid.Thus M=(-2 -4*(1/5), sqrt(2)*(1/5))= (-2 -4/5, sqrt(2)/5)= (-14/5, sqrt(2)/5)Compute |F2N|: distance from (2,0) to (-2, sqrt(2)) is sqrt( (-4)^2 + (sqrt(2))^2 )= sqrt(16 +2)=sqrt(18)=3sqrt(2)Compute |F1M|: distance from (-2,0) to (-14/5, sqrt(2)/5)= sqrt( (-14/5 +10/5)^2 + (sqrt(2)/5)^2 )= sqrt( (-4/5)^2 + 2/25 )= sqrt(16/25 +2/25)= sqrt(18/25)= (3sqrt(2))/5Thus, |F2N| - |F1M|=3sqrt(2) -3sqrt(2)/5= (15sqrt(2) -3sqrt(2))/5=12sqrt(2)/5≈3.39, which is greater than sqrt(6)≈2.45. Not matching.Hmm. This trial and error might not be effective. Let me try a different approach.Let me assume that the line F2N has slope m. Since N is on the ellipse above x-axis, coordinates of N can be found in terms of m, then compute M accordingly.Let me parametrize line F2N. Since it passes through F2(2,0) and has slope m, its equation is y = m(x -2).Intersection of this line with the ellipse gives point N.Substitute y = m(x -2) into the ellipse equation:x² /8 + [m²(x -2)^2]/4 =1Multiply both sides by 8:x² + 2m²(x -2)^2 =8Expand:x² +2m²(x² -4x +4)=8x² +2m²x² -8m²x +8m²=8(1 +2m²)x² -8m²x +8m² -8=0This is a quadratic in x. Solving this will give x-coordinates of intersection points. Since the line passes through F2(2,0), which is a focus, but the line might intersect the ellipse at another point N.Let me compute discriminant to find x.But since line passes through (2,0), substituting x=2, y=0 into the line equation satisfies, but (2,0) is a focus, not necessarily on the ellipse. Wait, the ellipse equation at x=2: 4/8 + y²/4 =0.5 + y²/4=1 ⇒ y²=2 ⇒ y=±√2. So (2,0) is not on the ellipse. Therefore, the line y = m(x -2) will intersect the ellipse at two points: one is N, and the other is not F2 (since F2 is not on the ellipse). So we can solve the quadratic equation.Let me write:(1 +2m²)x² -8m²x +8m² -8=0Let me denote a =1 +2m², b = -8m², c=8m² -8Then, solutions are x = [8m² ± sqrt(64m^4 -4*(1 +2m²)*(8m² -8))]/[2(1 +2m²)]Compute discriminant D:D = 64m^4 -4*(1 +2m²)*(8m² -8)Expand the second term:4*(1 +2m²)*(8m² -8) =4*[8m² -8 +16m^4 -16m²] =4*[ -8 +16m^4 -8m² ] =4*(-8 -8m² +16m^4) =-32 -32m² +64m^4Thus,D =64m^4 - (-32 -32m² +64m^4) =64m^4 +32 +32m² -64m^4 =32 +32m²=32(m² +1)Therefore, sqrt(D)=sqrt(32(m² +1))=4sqrt(2(m² +1))Thus, x = [8m² ±4sqrt(2(m² +1))]/[2(1 +2m²)] = [4m² ±2sqrt(2(m² +1))]/[1 +2m²]Therefore, x-coordinates of intersections are:x = [4m² +2sqrt(2(m² +1))]/(1 +2m²) and x = [4m² -2sqrt(2(m² +1))]/(1 +2m²)We need the point N above x-axis, so y =m(x -2) >0. Since m is the slope, if m >0, then x >2; if m <0, then x <2. But let's verify.Wait, if m >0, then y >0 requires x >2. If m <0, then y >0 requires x <2.But the ellipse's x ranges from -2√2≈-2.828 to 2√2≈2.828. So if m >0, then x >2 is possible up to 2.828. If m <0, x <2, possible down to -2.828.But let's find the coordinates.Case 1: m >0Then x >2, so take x = [4m² +2sqrt(2(m² +1))]/(1 +2m²). Let's compute this for m >0.Case 2: m <0Then x <2, take x = [4m² -2sqrt(2(m² +1))]/(1 +2m²). For m <0, need to check if this gives x <2.But this might get too involved. Alternatively, express N in terms of m and substitute into the other conditions.But recall that vec(F1M) is parallel to vec(F2N). If the line F2N has slope m, then vec(F2N) has slope m, so vec(F1M) also has slope m. Therefore, the line F1M also has slope m. Thus, point M lies on the line through F1(-2,0) with slope m: y = m(x +2)So M is the intersection of this line with the ellipse, other than F1.Therefore, similar to above, substitute y = m(x +2) into the ellipse equation:x² /8 + [m²(x +2)^2]/4 =1Multiply by 8:x² +2m²(x +2)^2=8Expand:x² +2m²(x² +4x +4)=8x² +2m²x² +8m²x +8m²=8(1 +2m²)x² +8m²x +8m² -8=0This quadratic in x has solutions. Since F1(-2,0) is on the line, x=-2 is a solution.Divide polynomial by (x +2):Using polynomial division or factor theorem.Let me plug x=-2 into the quadratic:(1 +2m²)(4) +8m²(-2) +8m² -8=4(1 +2m²) -16m² +8m² -8=4 +8m² -16m² +8m² -8=4 -8= -4≠0. Wait, this suggests x=-2 is not a root, but it should be because F1(-2,0) is on the line y=m(x +2) and the ellipse?Wait, substituting x=-2, y=0 into ellipse equation: (-2)^2/8 +0=4/8=0.5≠1. So F1(-2,0) is not on the ellipse. Wait, confusion again.The foci are at (-2,0) and (2,0), but these are not on the ellipse. The ellipse equation is x²/8 + y²/4=1. At x=-2, y²/4=1 -4/8=1 -0.5=0.5, so y=±sqrt(2). Therefore, the points (-2, ±sqrt(2)) are on the ellipse, but F1(-2,0) is not. Therefore, line y=m(x +2) passes through F1(-2,0) and intersects the ellipse at two points: one is M (above x-axis), the other is (-2, -sqrt(2)) if m=... Wait, but we need M above x-axis, so only one intersection?Wait, substituting y = m(x +2) into the ellipse:x²/8 + m²(x +2)^2/4=1Multiply by 8:x² + 2m²(x² +4x +4)=8x² +2m²x² +8m²x +8m²=8(1 +2m²)x² +8m²x +8m² -8=0This quadratic equation's solutions are x-coordinates of intersections. Since the line passes through (-2,0), which is not on the ellipse, both solutions are valid.But we need M above x-axis, so y =m(x +2) >0. So for m >0, x +2 >0 ⇒ x >-2. For m <0, x +2 <0 ⇒ x < -2.But given the ellipse's x ranges from -2√2≈-2.828 to 2.828. So for m >0: x >-2. But the line might intersect the ellipse at two points x1 >-2 and x2 >-2? Or one?Wait, solving the quadratic:x = [-8m² ± sqrt(64m^4 -4*(1 +2m²)*(8m² -8))]/[2*(1 +2m²)]Compute discriminant D:64m^4 -4*(1 +2m²)*(8m² -8) =64m^4 -4*(8m² -8 +16m^4 -16m²)= same as before.Wait, earlier when we computed for line F2N, discriminant was 32(m² +1). Similarly, here, perhaps.Compute D:=64m^4 -4*(1 +2m²)*(8m² -8)Expand:=64m^4 -4*[8m² -8 +16m^4 -16m²]=64m^4 -4*[ -8m² -8 +16m^4 ]=64m^4 +32m² +32 -64m^4=32m² +32=32(m² +1)Thus, sqrt(D)=sqrt(32(m² +1))=4sqrt(2(m² +1))Thus, x = [ -8m² ±4sqrt(2(m² +1)) ]/[2(1 +2m²)] = [ -4m² ±2sqrt(2(m² +1)) ]/(1 +2m²)Therefore, two solutions:x1 = [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)x2 = [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²)We need to determine which solution corresponds to M above x-axis.For m >0:y =m(x +2) >0 ⇒ x +2 >0 ⇒ x >-2Compute x1:x1 = [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)Whether this is greater than -2?Similarly, x2 is further left.So likely x1 is the valid solution.For m <0:y =m(x +2) >0 ⇒ m <0 and x +2 <0 ⇒ x < -2But x must be ≥-2√2≈-2.828. So possible.But this is getting very involved. Perhaps instead, relate the two conditions:From the problem, since vec(F1M) || vec(F2N), both lines have slope m. Thus, line F1M is y =m(x +2), line F2N is y=m(x -2). Their intersections with the ellipse give points M and N.But both points M and N are expressed in terms of m. Then, we can compute |F2N| and |F1M| in terms of m, set their difference to sqrt(6), and solve for m. Once m is found, the equation of line F2N is y =m(x -2), so we can write it.Let me attempt this.From previous calculations, for line F2N: y =m(x -2), intersection with ellipse gives point N with x-coordinate:x_N = [4m² ±2sqrt(2(m² +1))]/(1 +2m²)Similarly, for line F1M: y =m(x +2), intersection with ellipse gives point M with x-coordinate:x_M = [ -4m² ±2sqrt(2(m² +1)) ]/(1 +2m²)But we need to choose the correct sign in each case.For point N on F2N line, above x-axis:If m >0, then y_N >0 ⇒ x_N >2. From x_N expressions:x_N = [4m² +2sqrt(2(m² +1))]/(1 +2m²) or [4m² -2sqrt(2(m² +1))]/(1 +2m²)Since numerator of first solution: 4m² +2sqrt(2(m² +1)) is larger, and denominator 1 +2m² >0, so x_N first solution is larger. For m >0, we need x_N >2. Let's check if this solution satisfies.Compute x_N for m >0:Let me test m=1:x_N = [4*1 +2sqrt(2(1 +1))]/(1 +2*1) = [4 +2sqrt(4)]/3 = [4 +4]/3=8/3≈2.666>2. Valid.Second solution: [4 -2sqrt(4)]/3=(4-4)/3=0. But y=m(x -2) would be y=1*(0 -2)=-2<0. Not valid.Thus, for m >0, valid x_N is [4m² +2sqrt(2(m² +1))]/(1 +2m²)Similarly, for m <0, y_N >0 requires x_N <2. Let's take m=-1:x_N = [4*1 +2sqrt(2(1 +1))]/(1 +2*1)= same as above=8/3≈2.666>2. Which would give y= -1*(2.666 -2)= -0.666<0, invalid. So need the other solution:x_N = [4m² -2sqrt(2(m² +1))]/(1 +2m²)For m=-1:x_N = [4*1 -2sqrt(4)]/3= (4 -4)/3=0, then y=-1*(0 -2)=2>0. Valid.Thus, for m <0, valid x_N is [4m² -2sqrt(2(m² +1))]/(1 +2m²)Similarly, for point M:For m >0, line y=m(x +2) intersects ellipse at x_M:x_M = [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²) or x_M = [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²)Since m >0 and we need y_M >0, so x_M >-2.Compute x_M for m=1:x_M = [ -4 +2sqrt(4) ]/3= (-4 +4)/3=0. y=1*(0 +2)=2>0. Valid.Second solution: [ -4 -4 ]/3=-8/3≈-2.666. y=1*(-8/3 +2)= -2/3<0. Invalid.Thus, for m >0, valid x_M is [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)For m <0, need y_M >0, which requires x_M < -2. Compute x_M for m=-1:x_M = [ -4*1 +2sqrt(2(1 +1)) ]/(1 +2*1)= (-4 +4)/3=0. y=-1*(0 +2)=-2<0. Invalid.Second solution: [ -4*1 -2sqrt(4) ]/3= (-4 -4)/3=-8/3≈-2.666. y=-1*(-8/3 +2)= -1*(-2/3)=2/3>0. Valid.Thus, for m <0, valid x_M is [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²)Now, compute |F2N| and |F1M| for each case.Case 1: m >0Coordinates of N:x_N = [4m² +2sqrt(2(m² +1))]/(1 +2m²)y_N =m(x_N -2)Coordinates of M:x_M = [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)y_M =m(x_M +2)Compute |F2N|:distance from (2,0) to (x_N, y_N):sqrt( (x_N -2)^2 + y_N^2 ) = sqrt( (x_N -2)^2 + m²(x_N -2)^2 ) = |x_N -2| sqrt(1 +m²)Since x_N >2, this is (x_N -2) sqrt(1 +m²)Similarly, |F1M|:distance from (-2,0) to (x_M, y_M):sqrt( (x_M +2)^2 + y_M^2 ) = sqrt( (x_M +2)^2 + m²(x_M +2)^2 ) = |x_M +2| sqrt(1 +m²)Since x_M >-2 (for m >0), this is (x_M +2) sqrt(1 +m²)Given the condition |F2N| - |F1M| = sqrt(6):(x_N -2) sqrt(1 +m²) - (x_M +2) sqrt(1 +m²) = sqrt(6)Factor sqrt(1 +m²):[ (x_N -2) - (x_M +2) ] sqrt(1 +m²) = sqrt(6)Simplify inside:x_N -2 -x_M -2 = x_N -x_M -4Thus,(x_N -x_M -4) sqrt(1 +m²) = sqrt(6)Now, compute x_N -x_M:x_N -x_M = [4m² +2sqrt(2(m² +1))]/(1 +2m²) - [ -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)= [4m² +2sqrt(2(m² +1)) +4m² -2sqrt(2(m² +1)) ]/(1 +2m²)= (8m²)/(1 +2m²)Thus,x_N -x_M -4 = (8m²)/(1 +2m²) -4 = (8m² -4(1 +2m²))/(1 +2m²) = (8m² -4 -8m²)/(1 +2m²) = (-4)/(1 +2m²)Therefore,(-4)/(1 +2m²) * sqrt(1 +m²) = sqrt(6)Multiply both sides by (1 +2m²)/-4:sqrt(1 +m²) = - (1 +2m²) * sqrt(6)/4But sqrt(1 +m²) is positive, and right side is negative. Contradiction. Therefore, no solution for m >0.Case 2: m <0Coordinates of N:x_N = [4m² -2sqrt(2(m² +1))]/(1 +2m²)y_N =m(x_N -2)Coordinates of M:x_M = [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²)y_M =m(x_M +2)Compute |F2N|:distance from (2,0) to (x_N, y_N):sqrt( (x_N -2)^2 + y_N^2 ) = sqrt( (x_N -2)^2 + m²(x_N -2)^2 ) = |x_N -2| sqrt(1 +m²)Since m <0 and y_N >0, x_N <2, so x_N -2 <0, thus |x_N -2|=2 -x_N.Thus, |F2N|= (2 -x_N) sqrt(1 +m²)Similarly, |F1M|:distance from (-2,0) to (x_M, y_M):sqrt( (x_M +2)^2 + y_M^2 ) = sqrt( (x_M +2)^2 + m²(x_M +2)^2 ) = |x_M +2| sqrt(1 +m²)Since m <0 and y_M >0, x_M < -2, so x_M +2 <0, thus |x_M +2|= -x_M -2.Thus, |F1M|= (-x_M -2) sqrt(1 +m²)Condition |F2N| - |F1M| = sqrt(6):(2 -x_N) sqrt(1 +m²) - (-x_M -2) sqrt(1 +m²) = sqrt(6)Factor sqrt(1 +m²):[ (2 -x_N) +x_M +2 ] sqrt(1 +m²) = sqrt(6)Simplify inside:2 -x_N +x_M +2 =4 -x_N +x_MThus,(4 -x_N +x_M) sqrt(1 +m²) = sqrt(6)Now, compute x_M -x_N:x_M -x_N = [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²) - [4m² -2sqrt(2(m² +1)) ]/(1 +2m²)= [ -4m² -2sqrt(2(m² +1)) -4m² +2sqrt(2(m² +1)) ]/(1 +2m²)= (-8m²)/(1 +2m²)Therefore,4 -x_N +x_M =4 + (x_M -x_N) =4 -8m²/(1 +2m²)Thus,[4 -8m²/(1 +2m²)] sqrt(1 +m²) = sqrt(6)Let me combine the terms:[4(1 +2m²) -8m²]/(1 +2m²) * sqrt(1 +m²) = sqrt(6)Simplify numerator:4 +8m² -8m² =4Therefore,4/(1 +2m²) * sqrt(1 +m²) = sqrt(6)Multiply both sides by (1 +2m²)/4:sqrt(1 +m²) = sqrt(6) * (1 +2m²)/4Square both sides:1 +m² = (6)(1 +2m²)^2 /16Multiply both sides by 16:16(1 +m²) =6(1 +4m² +4m^4)Expand:16 +16m² =6 +24m² +24m^4Bring all terms to left:24m^4 +24m² +6 -16 -16m²=0Simplify:24m^4 +8m² -10=0Divide by 2:12m^4 +4m² -5=0Let me set u =m², then:12u² +4u -5=0Solve quadratic:u = [-4 ±sqrt(16 +240)]/(24) = [-4 ±sqrt(256)]/24 = [-4 ±16]/24Thus:u = (12)/24 = 0.5 or u = (-20)/24 =-5/6Since u =m² ≥0, discard negative solution. Thus, u=0.5.Thus, m²=0.5⇒m=±sqrt(0.5)=±√2/2. But in this case, m <0, so m=-√2/2.Thus, m=-√2/2.Therefore, the slope of line F2N is m=-√2/2. Hence, the equation of line F2N is:y -0 = m(x -2)y = (-√2/2)(x -2)Simplify:y = (-√2/2)x + √2Therefore, the equation is y = (-√2/2)x +√2.Let me verify this.First, compute coordinates of N.With m=-√2/2, compute x_N:x_N = [4m² -2sqrt(2(m² +1))]/(1 +2m²)Compute m²=0.5:x_N = [4*0.5 -2sqrt(2*(0.5 +1))]/(1 +2*0.5)= [2 -2sqrt(2*1.5)]/2= [2 -2sqrt(3)]/2= 1 -sqrt(3)Similarly, y_N =m(x_N -2)= (-√2/2)(1 -sqrt(3) -2)= (-√2/2)(-1 -sqrt(3))= (√2/2)(1 +sqrt(3))Check if N is on the ellipse:x_N²/8 + y_N²/4 = (1 -sqrt(3))²/8 + [ (√2/2(1 +sqrt(3)) )² ]/4Compute:(1 -2sqrt(3) +3)/8 + [ (2/4)(1 +2sqrt(3)+3) ]/4= (4 -2sqrt(3))/8 + [ (0.5)(4 +2sqrt(3)) ]/4= (0.5 -0.25sqrt(3)) + [ (2 +sqrt(3)) ]/4= 0.5 -0.25sqrt(3) +0.5 +0.25sqrt(3)= 1. Correct.Compute |F2N|:distance from (2,0) to (1 -sqrt(3), (√2/2)(1 +sqrt(3)))= sqrt( (1 -sqrt(3) -2)^2 + [ (√2/2)(1 +sqrt(3)) ]^2 )= sqrt( (-1 -sqrt(3))^2 + (2/4)(1 +2sqrt(3) +3) )= sqrt( (1 +2sqrt(3) +3) + (0.5)(4 +2sqrt(3)) )= sqrt(4 +2sqrt(3) +2 +sqrt(3))= sqrt(6 +3sqrt(3))Similarly, compute |F1M|:Coordinates of M:x_M = [ -4m² -2sqrt(2(m² +1)) ]/(1 +2m²)With m²=0.5:x_M = [ -4*0.5 -2sqrt(2*(0.5 +1)) ]/(1 +1)= [ -2 -2sqrt(3) ]/2= -1 -sqrt(3)y_M =m(x_M +2)= (-√2/2)(-1 -sqrt(3) +2)= (-√2/2)(1 -sqrt(3))= (√2/2)(sqrt(3) -1)Distance from (-2,0) to (-1 -sqrt(3), (√2/2)(sqrt(3) -1)):sqrt( (-1 -sqrt(3) +2)^2 + [ (√2/2)(sqrt(3) -1) ]^2 )= sqrt( (1 -sqrt(3))^2 + (2/4)(3 -2sqrt(3) +1) )= sqrt( (1 -2sqrt(3),<think>Okay, so I need to solve this problem about an ellipse. Let me start by understanding what is given and what I need to find. First, the ellipse is given by the equation x²/(2t²) + y²/t² = 1, where t > 0. The foci are F₁ and F₂, left and right foci respectively. The minimum distance from a point on the ellipse to F₂ is 2√2 - 2. Then there are points M and N on the ellipse above the x-axis, with vectors F₁M parallel to F₂N. There are three parts to the problem: find the equation of the ellipse, find the area of triangle F₁MN when a certain dot product is zero, and find the equation of the line F₂N when the difference in vector magnitudes is √6. Starting with part 1: Find the equation of ellipse C. I know that the standard form of an ellipse is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. The distance between the center and each focus is c, where c² = a² - b². In this problem, the ellipse equation is given as x²/(2t²) + y²/t² = 1. So comparing to the standard form, a² is 2t² and b² is t². Therefore, a = t√2 and b = t. Then, c² = a² - b² = 2t² - t² = t², so c = t. Therefore, the foci are located at (-c, 0) and (c, 0), which would be (-t, 0) and (t, 0) here. So F₁ is (-t, 0) and F₂ is (t, 0). The problem states that the minimum distance from a point on the ellipse to F₂ is 2√2 - 2. So I need to find the minimum distance from F₂ to any point on the ellipse and set that equal to 2√2 - 2. To find the minimum distance from a point on the ellipse to F₂, which is at (t, 0), we can parametrize the ellipse. Let me recall that parametric equations for an ellipse are x = a cosθ, y = b sinθ. So here, x = t√2 cosθ, y = t sinθ. The distance from a point (x, y) on the ellipse to F₂ (t, 0) is sqrt[(x - t)² + y²]. To find the minimum distance, we can minimize the square of the distance to avoid dealing with the square root. So let's compute D² = (x - t)² + y².Substituting x and y from the parametric equations:D² = (t√2 cosθ - t)² + (t sinθ)²= t²(√2 cosθ - 1)² + t² sin²θ= t²[(√2 cosθ - 1)² + sin²θ]Let me expand that:(√2 cosθ - 1)² + sin²θ = 2 cos²θ - 2√2 cosθ + 1 + sin²θ= (2 cos²θ + sin²θ) - 2√2 cosθ + 1But 2 cos²θ + sin²θ = cos²θ + (cos²θ + sin²θ) = cos²θ + 1So the expression becomes:(cos²θ + 1) - 2√2 cosθ + 1= cos²θ + 1 - 2√2 cosθ + 1= cos²θ - 2√2 cosθ + 2So D² = t² [cos²θ - 2√2 cosθ + 2]To minimize D², we can take the derivative with respect to θ and set it to zero. Let me compute the derivative:d(D²)/dθ = t² [2 cosθ (-sinθ) - 2√2 (-sinθ) + 0]= t² [-2 cosθ sinθ + 2√2 sinθ]Set derivative to zero:-2 cosθ sinθ + 2√2 sinθ = 0Factor out 2 sinθ:2 sinθ (-cosθ + √2) = 0So either sinθ = 0 or -cosθ + √2 = 0 → cosθ = √2. But cosθ can't be √2 because the maximum value of cosine is 1. So the critical points occur at sinθ = 0. Therefore, θ = 0 or π. But θ = 0 corresponds to the point (t√2, 0) on the ellipse, and θ = π corresponds to (-t√2, 0). Let's check these points.At θ = 0: the point is (t√2, 0). The distance to F₂ (t, 0) is |t√2 - t| = t(√2 - 1).At θ = π: the point is (-t√2, 0). The distance to F₂ (t, 0) is | - t√2 - t | = t(√2 + 1). But since we are looking for the minimum distance, the minimum occurs at θ = 0. Therefore, the minimum distance is t(√2 - 1). According to the problem, this is equal to 2√2 - 2. So:t(√2 - 1) = 2√2 - 2Factor out √2 - 1 on the right-hand side: 2(√2 - 1). Therefore:t(√2 - 1) = 2(√2 - 1)Divide both sides by (√2 - 1):t = 2So t is 2. Therefore, substituting back into the equation of the ellipse:x²/(2*(2)^2) + y²/(2)^2 = 1 → x²/(8) + y²/4 = 1Hence, the equation of the ellipse is x²/8 + y²/4 = 1.Wait, let me confirm that. Original equation was x²/(2t²) + y²/t² = 1. If t = 2, then 2t² = 2*4 = 8, and t² = 4. So yes, x²/8 + y²/4 = 1. That seems right.So part 1 answer is x²/8 + y²/4 = 1. Let me check again.Original minimum distance was given as 2√2 - 2. We calculated the minimum distance as t(√2 -1). Setting that equal to 2√2 -2. So t(√2 -1) = 2(√2 -1). Therefore, t=2. Then the equation is x²/(2*(2)^2) + y²/(2)^2 = x²/8 + y²/4 =1. Correct. So part 1 is done.Moving on to part 2: When the dot product F₁N · F₂N = 0, find the area of triangle F₁MN.First, let's note that F₁ and F₂ are the left and right foci. From part 1, t=2, so the foci are at (-2,0) and (2,0). So F₁=(-2,0), F₂=(2,0). Points M and N are on the ellipse above the x-axis. Also, vectors F₁M and F₂N are parallel. So vector F₁M is from F₁ to M, and vector F₂N is from F₂ to N. These vectors are parallel. So F₁M = k * F₂N for some scalar k. Also, since M and N are above the x-axis, their y-coordinates are positive.Given that the dot product F₁N · F₂N = 0. So vectors F₁N and F₂N are perpendicular. Wait, F₁N is the vector from F₁ to N, and F₂N is the vector from F₂ to N. Their dot product is zero.So, F₁N · F₂N = 0. Let's denote coordinates. Let N = (x, y), which is on the ellipse, so x²/8 + y²/4 =1, and y >0.Vector F₁N is (x - (-2), y - 0) = (x + 2, y)Vector F₂N is (x - 2, y - 0) = (x - 2, y)Their dot product is (x + 2)(x - 2) + y * y = x² -4 + y² = 0.So x² + y² =4.But N is on the ellipse x²/8 + y²/4 =1. So x² + 2y² =8. Therefore, we have two equations:x² + y² =4x² + 2y² =8Subtract the first equation from the second: (x² + 2y²) - (x² + y²) =8 -4 → y² =4. Therefore, y=2 (since y>0). Then substitute back into x² + y² =4: x² +4=4 →x²=0 →x=0.So point N is (0,2). Let me check: x²/8 + y²/4 =0 + 4/4=1. Correct. So N=(0,2). Now, since vector F₁M is parallel to vector F₂N. Let's find F₂N first. Vector F₂N is from F₂=(2,0) to N=(0,2): (0 -2, 2 -0)= (-2,2). So vector F₂N is (-2,2). Therefore, vector F₁M is parallel to (-2,2). So vector F₁M = k*(-2,2) for some scalar k. But vector F₁M is from F₁=(-2,0) to M=(x,y). So (x - (-2), y -0) = (x +2, y) = k*(-2,2). Therefore:x +2 = -2ky = 2kSo M has coordinates ( -2k -2, 2k )But M is on the ellipse x²/8 + y²/4 =1. Substituting:( (-2k -2)^2 )/8 + ( (2k)^2 )/4 =1Compute (-2k -2)^2 = [ -2(k +1) ]^2 =4(k +1)^2So 4(k +1)^2 /8 + 4k² /4 = ( (k +1)^2 )/2 + k² =1Multiply both sides by 2:(k +1)^2 + 2k² =2Expand (k +1)^2: k² +2k +1 +2k² =2Combine terms: 3k² +2k +1 =2Subtract 2: 3k² +2k -1 =0Solving quadratic equation: 3k² +2k -1=0Discriminant: 4 +12=16Solutions: k=(-2 ±4)/6So k= (-2 +4)/6=2/6=1/3Or k=(-2 -4)/6= -6/6= -1So k=1/3 or k=-1.But since points M and N are above the x-axis, and M is on the ellipse above the x-axis. So y-coordinate of M is 2k. Since y>0, 2k>0 ⇒k>0. Therefore, k=1/3.Thus, k=1/3. So coordinates of M are:x = -2*(1/3) -2 = -2/3 -2 = -8/3y=2*(1/3)=2/3Therefore, M=(-8/3, 2/3). Let's verify this is on the ellipse:x²/8 + y²/4 = (64/9)/8 + (4/9)/4 = (64/72) + (4/36)= (8/9) + (1/9)=9/9=1. Correct.So points M=(-8/3,2/3) and N=(0,2).Now, we need to find the area of triangle F₁MN. The coordinates of the three points are:F₁=(-2,0), M=(-8/3,2/3), N=(0,2)One way to find the area is using the shoelace formula.Arrange the coordinates:F₁: (-2, 0)M: (-8/3, 2/3)N: (0, 2)Back to F₁: (-2, 0)Compute the sum of products going one way:(-2)*(2/3) + (-8/3)*2 + 0*0 = (-4/3) + (-16/3) + 0 = (-20/3)Sum of products going the other way:0*(-8/3) + 2/3*0 + 2*(-2) =0 +0 -4= -4Take the absolute difference: |(-20/3) - (-4)| = | -20/3 +4 | = | -20/3 +12/3 | = | -8/3 | =8/3Area is half of that: 4/3.Alternatively, we can use vectors or determinant formula.Alternatively, compute vectors F₁M and F₁N, then take half the magnitude of the cross product.Vector F₁M: from F₁ to M: (-8/3 - (-2), 2/3 -0)= (-8/3 +6/3, 2/3)= (-2/3, 2/3)Vector F₁N: from F₁ to N: (0 - (-2), 2 -0)=(2,2)Cross product determinant:|i   j   k||-2/3 2/3 0||2   2   0|The cross product is (0,0, (-2/3)*2 - (2/3)*2)= (0,0, -4/3 -4/3)= (0,0, -8/3). The magnitude is 8/3. Half of that is 4/3. So area is 4/3.Therefore, the area of triangle F₁MN is 4/3.Wait, but let me check again using shoelace:Coordinates:F₁: (-2, 0)M: (-8/3, 2/3)N: (0, 2)Shoelace formula:Sum1 = (-2)*(2/3) + (-8/3)*2 + 0*0 = (-4/3) + (-16/3) + 0 = (-20/3)Sum2 = 0*(-8/3) + (2/3)*0 + 2*(-2) = 0 +0 -4 = -4Area = |Sum1 - Sum2| /2 = |(-20/3) - (-4)| /2 = |(-20/3 +12/3)| /2 = | -8/3 | /2 = (8/3)/2 =4/3. Correct.So the area is 4/3. Therefore, part 2 answer is 4/3.Moving on to part 3: When |F₂N| - |F₁M| =√6, find the equation of the line F₂N.First, let's recall that vectors F₁M and F₂N are parallel. So similar to part 2, but in this case, the difference in their magnitudes is √6. We need to find the equation of the line F₂N under this condition.First, let's parametrize points M and N. As before, since vectors F₁M and F₂N are parallel, we can denote vector F₁M = k * vector F₂N for some scalar k. However, since the vectors are parallel, the direction is the same or opposite. But since points M and N are above the x-axis, their y-coordinates are positive, so probably k is positive. Let me use similar notation as part 2.Let’s denote vector F₂N as (a, b). Then vector F₁M is parallel to this, so F₁M = k*(a, b). But also, points M and N are on the ellipse.Alternatively, let's use coordinates. Let’s denote point N as (x, y) on the ellipse above x-axis. Then vector F₂N is (x -2, y). Vector F₁M is parallel to this, so vector F₁M = k*(x -2, y). But vector F₁M is (M_x - (-2), M_y -0) = (M_x +2, M_y). So:M_x +2 = k(x -2)M_y = k yBut point M is on the ellipse: (M_x)^2 /8 + (M_y)^2 /4 =1So substituting:(k(x -2) -2)^2 /8 + (k y)^2 /4 =1But point N is also on the ellipse: x²/8 + y²/4 =1So we have two equations:1. (k(x -2) -2)^2 /8 + (k y)^2 /4 =12. x²/8 + y²/4 =1Also, the condition given is |F₂N| - |F₁M| =√6.Compute |F₂N| = sqrt( (x -2)^2 + y^2 )Compute |F₁M| = sqrt( (M_x +2)^2 + (M_y)^2 ) = sqrt( (k(x -2))^2 + (k y)^2 ) = k*sqrt( (x -2)^2 + y^2 ) = k |F₂N|Therefore, |F₂N| - |F₁M| = |F₂N| - k |F₂N| = |F₂N| (1 -k ) = √6So we have |F₂N| (1 -k ) = √6.But |F₂N| = sqrt( (x -2)^2 + y^2 )This seems complicated. Maybe there's a better way. Let's see.Alternatively, let's parametrize points M and N in terms of parameter k as in part 2. In part 2, we found that when the dot product was zero, k=1/3. Here, we need to find k such that |F₂N| - |F₁M|=√6.From part 2, we had:Vector F₁M = k*(vector F₂N)But in part 2, since vectors F₁M and F₂N are parallel, we can write vector F₁M = k*vector F₂N. Then |F₁M|=k*|F₂N|. So the condition |F₂N| - |F₁M| = (1 -k)|F₂N| =√6.So we need to find k and |F₂N| such that (1 -k)|F₂N|=√6. Also, points M and N are on the ellipse.Let me use the parametrization from part 2. Let me recall in part 2, we considered vector F₁M = k*(-2,2), leading to point M=(-2k -2, 2k). However, in that case, vector F₂N was (-2,2). But here, maybe vector F₂N can be any vector, not necessarily (-2,2). Wait, but in general, vector F₁M is parallel to vector F₂N, so we can express vector F₁M as k*(vector F₂N). But since they are parallel, the direction is same or opposite, so scalar multiple. Alternatively, let me parametrize point N as (x, y). Then vector F₂N is (x -2, y). Vector F₁M is parallel to this, so F₁M = λ*(x -2, y). But F₁M is also (M_x +2, M_y). Therefore:M_x +2 = λ(x -2)M_y = λ yAlso, point M is on the ellipse: (M_x)^2 /8 + (M_y)^2 /4 =1So substituting M_x = λ(x -2) -2 and M_y = λ y into the ellipse equation:[ (λ(x -2) -2 )^2 ] /8 + [ (λ y )^2 ] /4 =1But point N=(x, y) is on the ellipse, so x²/8 + y²/4 =1. So we have two equations:1. [ (λ(x -2) -2 )^2 ] /8 + [ (λ y )^2 ] /4 =12. x²/8 + y²/4 =1Additionally, |F₂N| - |F₁M| = √6. Let's compute |F₂N| = sqrt( (x -2)^2 + y^2 )|F₁M| = sqrt( (M_x +2)^2 + M_y^2 ) = sqrt( [λ(x -2)]^2 + [λ y]^2 ) = λ sqrt( (x -2)^2 + y^2 ) = λ |F₂N|Therefore, |F₂N| - λ |F₂N| = |F₂N| (1 - λ ) = √6So we have |F₂N| (1 - λ ) = √6. Let's denote |F₂N| = d. Then d(1 - λ ) =√6. So λ =1 - √6 /d.But this might not be straightforward. Maybe another approach.Let me try parametrizing point N using the parametric equations of the ellipse. Let θ be the parameter. So N=(2√2 cosθ, 2 sinθ). Because the ellipse equation is x²/8 + y²/4 =1, so semi-major axis a=2√2, semi-minor axis b=2. So parametric equations are x=2√2 cosθ, y=2 sinθ.Vector F₂N is (2√2 cosθ -2, 2 sinθ). Vector F₁M is parallel to this vector. So vector F₁M is k*(2√2 cosθ -2, 2 sinθ). But vector F₁M is (M_x +2, M_y). Therefore, M_x = k*(2√2 cosθ -2) -2, M_y= k*(2 sinθ). But point M is on the ellipse, so:[M_x]^2 /8 + [M_y]^2 /4 =1Substituting M_x and M_y:[ k*(2√2 cosθ -2) -2 ]^2 /8 + [ k*(2 sinθ) ]^2 /4 =1Let me expand this:First term:[ k*(2√2 cosθ -2) -2 ]^2 /8 = [ 2√2 k cosθ -2k -2 ]^2 /8= [ 2√2 k cosθ -2(k +1) ]^2 /8= [ 2(√2 k cosθ - (k +1)) ]^2 /8= 4[ √2 k cosθ - (k +1) ]^2 /8= [ √2 k cosθ - (k +1) ]^2 /2Second term:[ k*(2 sinθ) ]^2 /4 = (4k² sin²θ)/4 =k² sin²θTherefore, the equation becomes:[ √2 k cosθ - (k +1) ]² /2 +k² sin²θ =1Multiply through by 2 to eliminate denominator:[ √2 k cosθ - (k +1) ]² + 2k² sin²θ =2Expand the first square:(√2 k cosθ -k -1)^2 = ( -k -1 +√2 k cosθ )^2 = (√2 k cosθ - (k +1))^2= (√2 k cosθ)^2 - 2√2 k cosθ(k +1) + (k +1)^2=2k² cos²θ - 2√2 k(k +1) cosθ + (k +1)^2Thus, equation becomes:2k² cos²θ - 2√2 k(k +1) cosθ + (k +1)^2 +2k² sin²θ =2Combine terms:2k² (cos²θ + sin²θ) - 2√2 k(k +1) cosθ + (k +1)^2 =2Since cos²θ + sin²θ=1:2k² -2√2 k(k +1) cosθ + (k +1)^2 =2So:2k² + (k +1)^2 -2√2 k(k +1) cosθ =2Let’s compute 2k² + (k +1)^2:2k² +k² +2k +1=3k² +2k +1So equation becomes:3k² +2k +1 -2√2 k(k +1) cosθ =2Thus:3k² +2k +1 -2= 2√2 k(k +1) cosθSo:3k² +2k -1=2√2 k(k +1) cosθTherefore:cosθ= (3k² +2k -1)/(2√2 k(k +1))Now, also, the condition |F₂N| - |F₁M|=√6.Compute |F₂N|=sqrt( (2√2 cosθ -2)^2 + (2 sinθ)^2 )= sqrt( 8 cos²θ -8√2 cosθ +4 +4 sin²θ )= sqrt( 8 cos²θ +4 sin²θ -8√2 cosθ +4 )Factor:= sqrt(4(2 cos²θ + sin²θ) -8√2 cosθ +4 )But 2 cos²θ + sin²θ = cos²θ +1So:= sqrt(4(cos²θ +1) -8√2 cosθ +4 )= sqrt(4 cos²θ +4 -8√2 cosθ +4 )= sqrt(4 cos²θ -8√2 cosθ +8 )Factor 4 inside sqrt:= sqrt(4( cos²θ -2√2 cosθ +2 )) =2 sqrt( cos²θ -2√2 cosθ +2 )Similarly, |F₁M|= sqrt( (M_x +2)^2 + M_y^2 )But M_x =k*(2√2 cosθ -2) -2, so M_x +2 =k*(2√2 cosθ -2 )M_y=2k sinθTherefore, |F₁M|= sqrt( [k*(2√2 cosθ -2 )]^2 + [2k sinθ]^2 )= sqrt( k²(8 cos²θ -8√2 cosθ +4 ) +4k² sin²θ )= k sqrt(8 cos²θ -8√2 cosθ +4 +4 sin²θ )Again, 8 cos²θ +4 sin²θ=4(2 cos²θ + sin²θ)=4( cos²θ +1 )So:=k sqrt(4(cos²θ +1) -8√2 cosθ +4 )=k sqrt(4 cos²θ +4 -8√2 cosθ +4 )=k sqrt(4 cos²θ -8√2 cosθ +8 )= k*2 sqrt( cos²θ -2√2 cosθ +2 )Therefore, |F₁M|=2k sqrt( cos²θ -2√2 cosθ +2 )Therefore, |F₂N| - |F₁M|= 2 sqrt( cos²θ -2√2 cosθ +2 ) -2k sqrt( cos²θ -2√2 cosθ +2 )=2(1 -k) sqrt( cos²θ -2√2 cosθ +2 )Set equal to √6:2(1 -k) sqrt( cos²θ -2√2 cosθ +2 ) =√6Divide both sides by 2:(1 -k) sqrt( cos²θ -2√2 cosθ +2 )=√6 /2But from earlier, we had:cosθ= (3k² +2k -1)/(2√2 k(k +1))Therefore, cosθ is expressed in terms of k. Let's denote A= cosθ. Then:A= (3k² +2k -1)/(2√2 k(k +1))Also, sqrt( cos²θ -2√2 cosθ +2 ) can be rewritten. Let me compute the expression inside sqrt:cos²θ -2√2 cosθ +2Let me complete the square:cos²θ -2√2 cosθ + (√2)^2 +2 - (√2)^2 = (cosθ -√2)^2 +2 -2= (cosθ -√2)^2But wait, (cosθ -√2)^2=cos²θ -2√2 cosθ +2. So the expression inside sqrt is (cosθ -√2)^2. Therefore,sqrt( cos²θ -2√2 cosθ +2 )= |cosθ -√2|But since cosθ ≤1 and √2 ≈1.414>1, so cosθ -√2 ≤1 -1.414≈-0.414<0. Therefore, sqrt(...)=√2 -cosθ.Therefore, sqrt( cos²θ -2√2 cosθ +2 )=√2 -cosθ.Therefore, our previous equation becomes:(1 -k)(√2 -cosθ )=√6 /2But we have cosθ expressed in terms of k:cosθ= (3k² +2k -1)/(2√2 k(k +1))Therefore, substitute into the equation:(1 -k)(√2 - (3k² +2k -1)/(2√2 k(k +1)) ) =√6 /2Let me compute √2 - [ (3k² +2k -1)/(2√2 k(k +1)) ]Multiply numerator and denominator by √2:= [ 2√2 *k(k +1) - (3k² +2k -1) ] / [ 2√2 k(k +1) ]So:Numerator: 2√2 k(k +1) -3k² -2k +1Denominator:2√2 k(k +1)Therefore, expression becomes:(1 -k)*[2√2 k(k +1) -3k² -2k +1] / [2√2 k(k +1)] =√6 /2Multiply both sides by 2√2 k(k +1):(1 -k)[2√2 k(k +1) -3k² -2k +1 ] = √6 *√2 k(k +1)Simplify √6*√2=√12=2√3So right-hand side:2√3 k(k +1)Left-hand side: (1 -k)[2√2 k(k +1) -3k² -2k +1 ]Let me compute the terms inside the bracket:2√2 k(k +1) -3k² -2k +1First term:2√2 k(k +1)=2√2 k² +2√2 kThen subtract 3k² +2k and add 1:=2√2 k² +2√2 k -3k² -2k +1= (2√2 k² -3k²) + (2√2 k -2k) +1= k²(2√2 -3) + k(2√2 -2) +1Therefore, left-hand side:(1 -k)[k²(2√2 -3) +k(2√2 -2) +1 ]Therefore, the equation is:(1 -k)[k²(2√2 -3) +k(2√2 -2) +1 ] =2√3 k(k +1)This seems very complicated. Perhaps there is a better approach. Let me think.Alternatively, since we have two equations:1. 3k² +2k -1=2√2 k(k +1) cosθ (from earlier)2. (1 -k)(√2 -cosθ )=√6 /2We can substitute cosθ from equation 1 into equation 2.From equation 1: cosθ= (3k² +2k -1)/(2√2 k(k +1))Substitute into equation 2:(1 -k)(√2 - (3k² +2k -1)/(2√2 k(k +1)) )=√6 /2This is the same as what I had earlier. Let me try to compute this step by step.First, compute √2 - [ (3k² +2k -1)/(2√2 k(k +1)) ]Let’s write √2 as (2√2 k(k +1))/(2√2 k(k +1)) to have the same denominator:√2 = [2√2 k(k +1) *√2 ] / [2√2 k(k +1) ]Wait, no, better to compute:√2 - (3k² +2k -1)/(2√2 k(k +1)) = [2√2 k(k +1) *√2 - (3k² +2k -1) ] / [2√2 k(k +1)]Wait, that might not help. Let me compute numerator:Let’s multiply √2 by denominator 2√2 k(k +1):Numerator: √2 *2√2 k(k +1) - (3k² +2k -1)=2*2 k(k +1) - (3k² +2k -1)=4k(k +1) -3k² -2k +1=4k² +4k -3k² -2k +1= (4k² -3k²) + (4k -2k) +1=k² +2k +1=(k +1)^2Therefore,√2 - (3k² +2k -1)/(2√2 k(k +1)) = (k +1)^2 / [2√2 k(k +1) ]= (k +1)/(2√2 k )Therefore, equation 2 becomes:(1 -k)*(k +1)/(2√2 k )=√6 /2Multiply both sides by 2√2 k:(1 -k)(k +1)=√6 *√2 kSimplify left-hand side: (1 -k)(1 +k)=1 -k²Right-hand side: √12 k=2√3 kSo:1 -k²=2√3 kRearranged:k² +2√3 k -1=0Solve quadratic equation:k= [ -2√3 ±sqrt( (2√3)^2 +4 ) ]/2= [ -2√3 ±sqrt(12 +4) ]/2= [ -2√3 ±sqrt(16) ]/2= [ -2√3 ±4 ]/2= [ -√3 ±2 ]So k= -√3 +2 or k= -√3 -2But since in our previous parametrization, point M is above x-axis, so M_y=2k sinθ>0. Since θ is such that N is above x-axis, sinθ>0, so 2k sinθ>0 implies k>0.So check the solutions:k= -√3 +2≈ -1.732+2≈0.268>0k= -√3 -2≈-3.732<0. Discard.Thus, k=2 -√3.Therefore, k=2 -√3. Now, we can find cosθ from equation 1:cosθ= (3k² +2k -1)/(2√2 k(k +1))Compute k=2 -√3, so first compute k +1=3 -√3Compute k²= (2 -√3)^2=4 -4√3 +3=7 -4√3Compute numerator:3k² +2k -1=3*(7 -4√3) +2*(2 -√3) -1=21 -12√3 +4 -2√3 -1=21 +4 -1 -12√3 -2√3=24 -14√3Denominator:2√2 k(k +1)=2√2*(2 -√3)(3 -√3)First compute (2 -√3)(3 -√3):=6 -2√3 -3√3 + (√3)^2=6 -5√3 +3=9 -5√3Thus denominator=2√2*(9 -5√3 )Therefore, cosθ= (24 -14√3)/(2√2*(9 -5√3 ))Simplify numerator and denominator:Factor numerator:2*(12 -7√3 )Denominator:2√2*(9 -5√3 )Cancel 2:cosθ= (12 -7√3 )/(√2*(9 -5√3 ))Rationalize denominator:Multiply numerator and denominator by (9 +5√3 ):cosθ= [ (12 -7√3)(9 +5√3 ) ] / [√2*(9 -5√3)(9 +5√3 ) ]Compute denominator first:(9)^2 - (5√3)^2=81 -75=6So denominator=√2*6=6√2Numerator:(12)(9) +12*5√3 -7√3*9 -7√3*5√3=108 +60√3 -63√3 -35*3=108 + (60√3 -63√3) -105=108 -105 -3√3=3 -3√3Thus, cosθ= (3 -3√3 )/(6√2 )= 3(1 -√3 )/(6√2 )= (1 -√3 )/(2√2 )Rationalize the denominator:Multiply numerator and denominator by √2:= (1 -√3 )√2 / (2*2)= (√2 -√6 )/4Therefore, cosθ= (√2 -√6 )/4Now, we can find sinθ using sin²θ=1 -cos²θCompute cosθ=(√2 -√6)/4. Let's square it:cos²θ= ( (√2 -√6 )² ) /16 = (2 -2√12 +6)/16= (8 -4√3)/16= (2 -√3)/4Thus, sin²θ=1 - (2 -√3)/4= (4 -2 +√3)/4=(2 +√3)/4Since sinθ>0, sinθ=√(2 +√3)/2. Let me simplify √(2 +√3):Recall that √(2 +√3)= (√3 +1)/√2. Let me check:(√3 +1)/√2 squared= (3 +2√3 +1)/2= (4 +2√3)/2=2 +√3. Correct.Therefore, sinθ= (√3 +1)/(2√2 )Therefore, coordinates of N are:x=2√2 cosθ=2√2*(√2 -√6)/4=2√2*(√2 -√6)/4= [2*2 -2√12 ]/4= [4 -4√3 ]/4=1 -√3y=2 sinθ=2*(√3 +1)/(2√2 )=(√3 +1)/√2= (√6 +√2)/2Therefore, point N=(1 -√3, (√6 +√2)/2 )Therefore, the line F₂N passes through F₂=(2,0) and N=(1 -√3, (√6 +√2)/2 )Compute the slope of F₂N:Slope m= [ ( (√6 +√2)/2 -0 ) / (1 -√3 -2 ) ]= [ (√6 +√2)/2 ] / ( -1 -√3 )= (√6 +√2)/2 divided by (-1 -√3 )Multiply numerator and denominator by -1:= (√6 +√2)/2 divided by (1 +√3 )Multiply numerator and denominator by (1 -√3 ):= (√6 +√2)(1 -√3 ) / [2*(1 -3) ]= (√6 +√2)(1 -√3 ) / [2*(-2) ]= -(√6 +√2)(1 -√3 ) /4Let me compute (√6 +√2)(1 -√3 ):=√6*1 -√6*√3 +√2*1 -√2*√3=√6 -√18 +√2 -√6Simplify:√6 -3√2 +√2 -√6= (√6 -√6 ) + (-3√2 +√2 )= -2√2Therefore, numerator= -2√2Thus, slope m= -(-2√2 ) /4=2√2 /4=√2 /2Therefore, slope is √2 /2.Hence, the equation of line F₂N is y -0= (√2 /2)(x -2 )Simplify: y= (√2 /2)x -√2 Alternatively, multiply both sides by 2:√2 x -2y -2√2=0But perhaps the answer is better written in simpler form. So:y= (√2 /2)x -√2 Or rationalizing, but it's fine as is. Alternatively, to write in standard form: √2 x -2y -2√2 =0But maybe rationalizing the denominator:Multiply numerator and denominator by √2:y= ( (√2 /2 )x -√2 )But maybe leave as is. Alternatively, write as y = (√2/2)(x - 2)Thus, the equation is y= (√2/2)x -√2.Let me verify if this line passes through F₂=(2,0):Plugging x=2: y= (√2/2)*2 -√2=√2 -√2=0. Correct.Also passes through N=(1 -√3, (√6 +√2)/2 )Compute y-coordinate using x=1 -√3:(√2/2)(1 -√3 ) -√2= (√2/2 - (√6)/2 ) -√2= -√2/2 -√6/2= -(√2 +√6)/2. Wait, but that's not matching. Did I make a mistake?Wait, the y-coordinate of N is (√6 +√2)/2. Let me check:Plug x=1 -√3 into the line equation:y= (√2/2)(1 -√3 ) -√2 = (√2/2 - (√6)/2 ) -√2 = (√2/2 -√2 ) -√6/2 = (-√2/2 ) -√6/2 = - (√2 +√6 ) / 2But this contradicts the y-coordinate of N which is (√6 +√2)/2. So there must be an error in calculation.Wait, where did I make a mistake?Wait, previously, in computing the slope:I had:Slope m= [ ( (√6 +√2)/2 -0 ) / (1 -√3 -2 ) ]= [ (√6 +√2)/2 ] / ( -1 -√3 )But then I multiplied numerator and denominator by -1:= [ (√6 +√2)/2 ] / ( -1 -√3 ) = - [ (√6 +√2)/2 ] / (1 +√3 )But when multiplying numerator and denominator by -1, it's equivalent to multiplying by (-1)/(-1)=1, so the value doesn't change. But actually, the original fraction is:[ (√6 +√2)/2 ] / ( -1 -√3 ) = - [ (√6 +√2)/2 ] / (1 +√3 )Then, multiplying numerator and denominator by (1 -√3 ):Numerator: - [ (√6 +√2)/2 ]*(1 -√3 )Denominator: (1 +√3 )(1 -√3 )=1 -3= -2Therefore, the slope m= - [ (√6 +√2 )(1 -√3 )/2 ] / (-2 )= [ (√6 +√2 )(1 -√3 ) ] /4Wait, previously I thought it simplifies to √2 /2, but apparently not. Let me recalculate:Let me recompute the slope:Numerator of slope: y_N - y_F2= (√6 +√2)/2 -0=(√6 +√2)/2Denominator: x_N -x_F2= (1 -√3 ) -2= -1 -√3 Therefore, slope m= (√6 +√2)/2 divided by (-1 -√3 )Multiply numerator and denominator by (-1 +√3 ) to rationalize denominator:m= [ (√6 +√2)/2 * (-1 +√3 ) ] / [ (-1 -√3 )(-1 +√3 ) ]Denominator: (-1)^2 - (√3 )^2=1 -3=-2Numerator: (√6 +√2)(-1 +√3 )/2Let me compute numerator:(√6 +√2)(-1 +√3 )= -√6 +√18 -√2 +√6Simplify:-√6 +√18 -√2 +√6= (-√6 +√6 ) + (√18 -√2 )=0 + (3√2 -√2 )=2√2Therefore, numerator=2√2 /2=√2Denominator=-2Therefore, slope m=√2 / (-2 )= -√2 /2Ah! So slope is -√2 /2, not √2 /2 as I previously thought. So there was a mistake in sign.Therefore, the correct slope is -√2 /2. Therefore, equation of the line is:y -0= -√2 /2 (x -2 )Thus, y= - (√2 /2 )x +√2 Now check if this passes through N=(1 -√3, (√6 +√2 )/2 )Compute y when x=1 -√3:y= - (√2 /2 )(1 -√3 ) +√2= -√2 /2 + (√6 )/2 +√2= (√2 /2 ) + (√6 )/2= (√2 +√6 )/2. Which matches the y-coordinate of N. Correct.Therefore, the equation of line F₂N is y= - (√2 /2 )x +√2 We can write this as:y= - (√2/2)x +√2 Alternatively, multiply both sides by 2 to eliminate the fraction:2y= -√2 x +2√2 Bring all terms to one side:√2 x +2y -2√2 =0But it's customary to rationalize or present in a simplified form. Alternatively, write in standard form Ax + By + C=0:√2 x +2y -2√2 =0Or if we prefer, divide by √2:x + √2 y -2=0But this might not be necessary. Alternatively, keep it as y= - (√2 /2 )x +√2 But in the problem statement, it says "find the equation of the line F₂N". The answer can be expressed in various forms, but slope-intercept form is acceptable. Alternatively, they might prefer standard form. Let me check.Given the answer is likely to be in a simplified radical form, maybe written as √2 x + 2y = 2√2. Let me verify:Starting from y= - (√2 /2 )x +√2 Multiply both sides by 2:2y= -√2 x +2√2 Bring √2 x to the left:√2 x +2y=2√2 Yes, this is correct. So the equation is √2 x +2y=2√2. Alternatively, divide both sides by √2:x + √2 y=2√2 /√2=2. But then x +√2 y=2. Wait, but let me check:From equation √2 x +2y=2√2. Divide both sides by √2:x + (2/√2 )y=2√2 /√2=2Simplify 2/√2=√2. So:x +√2 y=2. Yes, this is also correct.So the equation can be written as x +√2 y=2.Therefore, the equation of line F₂N is x +√2 y=2.Alternatively, in different forms. Both are correct. Depending on the desired format.But let me check with coordinates of F₂ and N:F₂=(2,0): 2 +√2*0=2. Correct.N=(1 -√3, (√6 +√2 )/2 )Check x +√2 y:1 -√3 +√2*( (√6 +√2 )/2 )=1 -√3 + ( √2*√6 +√2*√2 )/2=1 -√3 + (√12 +2 )/2=1 -√3 + (2√3 +2 )/2=1 -√3 +√3 +1=2. Correct.So equation x +√2 y=2 passes through both F₂ and N.Therefore, the equation of the line F₂N is x +√2 y=2.Final Answer1. The equation of ellipse ( C ) is boxed{dfrac{x^2}{8} + dfrac{y^2}{4} = 1}.2. The area of triangle ( F_1MN ) is boxed{dfrac{4}{3}}.3. The equation of the line ( F_2N ) is boxed{x + sqrt{2} y = 2}.</think>Given the points ( F_1 ) and ( F_2 ) are respectively the left and right foci of the ellipse ( C: dfrac{x^2}{2t^2} + dfrac{y^2}{t^2} = 1 ) ((t > 0)), and the minimum distance from a point on ellipse ( C ) to the point ( F_2 ) is ( 2sqrt{2} - 2 ). Points ( M ) and ( N ) are located above the x-axis on the ellipse ( C ), and the vector (vec{F_1M}) is parallel to the vector (vec{F_2N}).1. Find the equation of ellipse ( C ): The given ellipse equation is (dfrac{x^2}{2t^2} + dfrac{y^2}{t^2} = 1). The semi-major axis (a = tsqrt{2}) and the semi-minor axis (b = t). The distance between the center and each focus (c = t). The minimum distance from a point on the ellipse to (F_2) is (t(sqrt{2} - 1)). Setting this equal to (2sqrt{2} - 2), we find (t = 2). Substituting (t = 2) into the ellipse equation, we get: [ dfrac{x^2}{8} + dfrac{y^2}{4} = 1 ] Thus, the equation of the ellipse is (boxed{dfrac{x^2}{8} + dfrac{y^2}{4} = 1}).2. When the dot product (vec{F_1N} cdot vec{F_2N} = 0), find the area of triangle (F_1MN): Given the dot product (vec{F_1N} cdot vec{F_2N} = 0), we find (N = (0, 2)). The vector (vec{F_2N} = (-2, 2)), leading to (M = left(-dfrac{8}{3}, dfrac{2}{3}right)). Using the shoelace formula for the area of triangle (F_1MN) with vertices (F_1(-2, 0)), (Mleft(-dfrac{8}{3}, dfrac{2}{3}right)), and (N(0, 2)), we find the area to be (dfrac{4}{3}). Thus, the area is (boxed{dfrac{4}{3}}).3. When (|vec{F_2N}| - |vec{F_1M}| = sqrt{6}), find the equation of the line (F_2N): Given (|vec{F_2N}| - |vec{F_1M}| = sqrt{6}), we find the coordinates of (N) as ((1 - sqrt{3}, dfrac{sqrt{6} + sqrt{2}}{2})). The line (F_2N) passes through (F_2(2, 0)) and (N(1 - sqrt{3}, dfrac{sqrt{6} + sqrt{2}}{2})). The slope of the line is (-dfrac{sqrt{2}}{2}), leading to the equation (x + sqrt{2}y = 2). Thus, the equation of the line is (boxed{x + sqrt{2} y = 2}).]

🔑:1. As points F_1 and F_2 are the foci of the ellipse C: dfrac{x^2}{2t^2} + dfrac{y^2}{t^2} = 1, we have that a = sqrt{2}t text{ and } c = t. Given the minimum distance from a point on ellipse C to point F_2 is 2sqrt{2} - 2, the distance between the vertices and foci is:a - c = sqrt{2}t - t = 2sqrt{2} - 2. Solving for t yields t = 2. Therefore, the equation of the ellipse is:boxed{dfrac{x^2}{8} + dfrac{y^2}{4} = 1}.2. From part (1), we have F_1(-2,0) and F_2(2,0). We can parametrize point N on ellipse C as N(2sqrt{2}costheta, 2sintheta). Thus, we have:vec{F_1N} = (2sqrt{2}costheta + 2, 2sintheta), vec{F_2N} = (2sqrt{2}costheta - 2, 2sintheta).Given vec{F_1N} cdot vec{F_2N} = 0, we solve the equation:(2sqrt{2}costheta + 2)(2sqrt{2}costheta - 2) + 4sin^2theta = 0.This yields costheta = 0 and sintheta = 1. Hence, N(0,2) and vec{F_2N}=(-2,2), which implies that the slope k_{F_2N} = 1. Since vector vec{F_1M} is parallel to vector vec{F_2N}, the slope of line F_1M is also -1. Thus, the line equation of F_1M is y = -x - 2. Solving this system of equations:begin{cases}y = -x - 2 dfrac{x^2}{8} + dfrac{y^2}{4} = 1end{cases}yields x = 0, y = -2 (discarded), or x = -dfrac{8}{3}, y = dfrac{2}{3}. Therefore, Mleft(-dfrac{8}{3}, dfrac{2}{3}right) and |F_1M| = dfrac{2sqrt{2}}{3}. The distance of point N to the line y = -x - 2 is d = 2sqrt{2}. Consequently, the area of triangle F_1MN is:boxed{S_{F_1MN} = dfrac{1}{2}|F_1M|cdot d = dfrac{1}{2} cdot dfrac{2sqrt{2}}{3} cdot 2sqrt{2} = dfrac{4}{3}}.3. Since vectors vec{F_1M} and vec{F_2N} are parallel, we can write lambda vec{F_1M} = vec{F_2N}. Given |vec{F_2N}| - |vec{F_1M}| = sqrt{6}, it follows that lambda > 1, and we can establish:lambda(x_1 + 2) = x_2 - 2, ; y_2 = lambda y_1, ; (x_2, y_2), (x_1, y_1) in C.Substituting these into the ellipse equation, x_2^2 + 2y_2^2 = 8, and solving for x_1 yields:x_1 = dfrac{1-3lambda}{lambda} = dfrac{1}{lambda} - 3.Consequently, we get y_1^2 = 4 - dfrac{(1-3λ)^2}{2lambda}, and hence |vec{F_1M}| = dfrac{λ+1}{sqrt{2}λ}. Solving the equation (λ-1) cdot dfrac{λ+1}{sqrt{2}λ} = sqrt{6} gives us lambda = 2 + sqrt{3} (the other root is discarded). Therefore, x_1 = -1 - sqrt{3}, and y_1 = dfrac{sqrt{3}-1}{sqrt{2}}. Thus, the slope k_{F_1M} = -dfrac{sqrt{2}}{2}, and the equation of the line F_2N is y = -dfrac{sqrt{2}}{2}(x - 2). Simplifying, we have the final line equation:boxed{x + sqrt{2}y - 2 = 0}.