JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the ellipse \(\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}\gt\mathrm{b}\) and \(\mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1, \mathrm{~A} \lt \mathrm{B}\) have same eccentricity \(\frac{1}{\sqrt{3}}\). Let the product of their lengths of latus rectums be \(\frac{32}{\sqrt{3}}\), and the distance between the foci of \(E_1\) be 4. If \(E_1\) and \(E_2\) meet at \(A, B, C\) and \(D\), then the area of the quadrilateral \(A B C D\) equals :
- A \(\frac{12 \sqrt{6}}{5}\)
- B \(6 \sqrt{6}\)
- C \(\frac{18 \sqrt{6}}{5}\)
- D \(\frac{24 \sqrt{6}}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{24 \sqrt{6}}{5}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & 2 a e=4 \\ & \Rightarrow \quad a=2 \sqrt{3} \\ & \Rightarrow \quad 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8 \\ & \\ & \frac{2 b^2}{a} \times \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \\ & \Rightarrow \\ & \frac{2 \times 8}{2 \sqrt{3}} \times \frac{2…
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