Let the ellipse ( E _ 1 : x^2 a^2 + y^2 ~b^2 =1, a ) and (E _ 2 : x^2 ~A^2 + y^2 ~B^2 = 1, ~A B ) have same eccentricity ( 1 3 ). Let the product of their lengths of latus rectums be ( 32 3 ), and the distance between

Correct answer: D. ( 24 6 5 ) aligned & 2 a e=4 & a=2 3 & 1-(b^2)/(12)=(1)/(3) b^2=8 & & (2 b^2)/(a) × (2 A^2)/(B)= 32 3 & & 2 × 8 2 3 × (2 A^2)/(B)= 32 3 & & (A^2)/(B)=2 A^2=2 B & & 1-(A^2)/(B)=(1)/(3) & & B=3 A^2=6 aligned E_1: (x^2)/(12)+(y^2)/(8)=1 ....(i) E_1: (x^2)/(6)+(y^2)/(9)=1 ...(ii) On solving (i) \& (ii) aligned &(x, y)=( 6 5 , 6 5 ),( - 6 5 , 6 5 ),( 6 5 , -6 5 ), &( - 6 5 , -6 5 ) aligned Four points are vertices of rectangle area = 24 6 5

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Ellipse

Let the ellipse ( E _ 1 : x^2 a^2 + y^2 ~b^2 =1, a ) and (E _ 2 :...

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}=

\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 :

Options

A(\frac{12 \sqrt{6}}{5})
B(6 \sqrt{6})
C(\frac{18 \sqrt{6}}{5})
D(\frac{24 \sqrt{6}}{5})

Correct answer

D. (\frac{24 \sqrt{6}}{5})

Step-by-step solution

\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 A^2}{B}=\frac{32}{\sqrt{3}} \\ & \Rightarrow \\ & \frac{A^2}{B}=2 \Rightarrow A^2=2 B \\ & \\ & \quad 1-\frac{A^2}{B}=\frac{1}{3} \\ & \Rightarrow \\ & B=3 \Rightarrow A^2=6\end{aligned}

E_1: \frac{x^2}{12}+\frac{y^2}{8}=1

....(i)

E_1: \frac{x^2}{6}+\frac{y^2}{9}=1

...(ii) On solving (i) \& (ii)

\begin{aligned} &(x, y)=\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)\end{aligned}

Four points are vertices of rectangle area

=\frac{24 \sqrt{6}}{5}

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