Let the ellipse , a > b and , A < B have same eccentricity . Let the product of their lengths of latus rectums be , and the distance between the foci of be 4. If and meet at A, B, C and D, then the area of the quadrilateral ABCD equals : [2025

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Solution

Q.
Let the ellipse $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $E_{2} : \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ , A < 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 ABCD equals : [2025]

1 $6 \sqrt{6}$

2 $\frac{12 \sqrt{6}}{5}$

3 $\frac{24 \sqrt{6}}{5}$

4 $\frac{18 \sqrt{6}}{5}$

Ans.
(3)

We have, $2 a e = 4 \Rightarrow 2 a (\frac{1}{\sqrt{3}}) = 4 \Rightarrow a = 2 \sqrt{3}$

$b^{2} = a^{2} (1 - e^{2}) = 12 (1 - \frac{1}{3}) = 8$

Now, $\Rightarrow \frac{2 b^{2}}{a} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}}$

$\Rightarrow \frac{2 \cdot 8}{2 \sqrt{3}} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}} \Rightarrow A^{2} = 2 B$

Alao, $\Rightarrow A^{2} = B^{2} (1 - e^{2}) \Rightarrow 2 B = B^{2} \cdot \frac{2}{3} \Rightarrow B = 3 \Rightarrow A^{2} = 6$

$∴ E_{1} : \frac{x^{2}}{12} + \frac{y^{2}}{8} = 1$ ... (i)

and $E_{2} : \frac{x^{2}}{6} + \frac{y^{2}}{9} = 1$ ... (ii)

Solving (i) and (ii), we get

$A (\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), B (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), C (\frac{\sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}}), D (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}})$

which form a rectangle.

$∴$ Required area $\frac{12}{\sqrt{5}} \times \frac{2 \sqrt{6}}{\sqrt{5}} = \frac{24 \sqrt{6}}{5}$

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