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JEE Main 29-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2025 Paper - Shift 14 Marks
MCQ
Let the ellipse, $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~B}^{2}}=1, \mathrm{~A}<\mathrm{B}$ 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 ABCD equals:
  • A
    $6 \sqrt{6}$
  • B
    $\frac{18 \sqrt{6}}{5}$
  • C
    $\frac{12 \sqrt{6}}{5}$
  • D
    $\frac{24 \sqrt{6}}{5}$
✓

Answer

(D)
Sol. $\quad 2 \mathrm{ae}=4$
$2 \mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4$
$\Rightarrow \mathrm{a}=2 \sqrt{3}$
$\Rightarrow 1-\frac{\mathrm{b}^{2}}{12}=\frac{1}{3} \Rightarrow \mathrm{~b}^{2}=8$
Now $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}} \cdot \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^{2}=2 B$
$1-\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}=\frac{1}{3} \Rightarrow 1-\frac{2 \mathrm{~B}}{\mathrm{~B}^{2}}=\frac{1}{3} \Rightarrow \mathrm{~B}=3$
$\Rightarrow A^{2}=6$
$\frac{x^{2}}{12}+\frac{y^{2}}{8}=1$ ...(1)
$\frac{x^{2}}{6}+\frac{y^{2}}{9}=1$ ...(2)
On solving (1) & (2) we get
$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)$
The four points are vertices of rectangle and its area $=$ $\frac{24 \sqrt{6}}{5}$

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