Let the ellipse E1:a2x2+b2y2=1,a>b and E2:A2x2+B2y2=1,A<B have the same eccentricity 31. Let the product of their lengths of latus rectums be 332, and the distance between the foci of E1 be 4. If E1 and E2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals:
Get the complete, step-by-step math solution for: "Let the ellipse {E}_1: (x²)/(a²) + (y²)/(b²) = 1, {a} > {b} and {E}_2: (x²)/(A²) + (y²)/(B²) = 1, {A} < {B} have the same eccentricity {1}{√(3)}. Let ...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Determine parameters for Ellipse E1
We are given that the eccentricity of ellipse E1 is e=31 and the distance between its foci is 2ae=4. We can use these to find the values of a and b for E1. Recall that for an ellipse with a>b, b2=a2(1−e2).
Step 2: Calculate 'a' and 'b' for E1
From 2ae=4 and e=31, we find a=2e4=1/32=23. Then, using b2=a2(1−e2), we calculate b2=(23)2(1−(1/3)2)=12(1−1/3)=12(2/3)=8. So, b=8=22.
Step 3: Determine parameters for Ellipse E2
For ellipse E2, since A < B, the major axis is along the y -axis. Its eccentricity is e′=31, and A2=B2(1−e′2). The length of the latus rectum for E1 is L1=a2b2 and for E2 is L2=B2A2. We are given that L1L2=332.
Step 4: Calculate 'A' and 'B' for E2
Substitute the values of a and b2 for E1 into the product of latus rectums equation: B2A2⋅232(8)=332. This simplifies to B316A2=332, which gives A2=2B. Also, for E2, A2=B2(1−e′2)=B2(1−(1/3)2)=B2(2/3). Equating the two expressions for A2: 2B=B2(2/3)⟹B=3. Then A2=2(3)=6, so A=6.
Step 5: Find intersection points and area of quadrilateral ABCD
The equations of the ellipses are E1:12x2+8y2=1 and E2:6x2+9y2=1. To find the intersection points, we solve these equations simultaneously. Multiply the second equation by 2: 3x2+92y2=2. Subtract this from the first equation: (12x2−3x2)+(8y2−92y2)=1−2. This gives −123x2+729y2−16y2=−1⟹−4x2−727y2=−1⟹4x2+727y2=1. This is incorrect. Let's solve for x2 and y2 directly. From E1, x2=12(1−y2/8). From E2, x2=6(1−y2/9). Equating them: 12(1−y2/8)=6(1−y2/9)⟹2(1−y2/8)=1−y2/9⟹2−y2/4=1−y2/9⟹1=y2/4−y2/9=369y2−4y2=365y2. So y2=536⟹y=±56. Substitute y2 back into x2=6(1−y2/9)=6(1−5⋅936)=6(1−54)=6(51)=56. So x=±56. The intersection points are (±56,±56). These points form a rectangle with vertices (x0,y0),(−x0,y0),(−x0,−y0),(x0,−y0). The area of this rectangle is 2∣x0∣⋅2∣y0∣=4∣x0y0∣. Area =4⋅56⋅56=4⋅566=5246.