Ellipse — JEE Main · 2025 · 29 Jan 2025 · Shift 1 Question with Solution

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}{3}

. Let the product of their lengths of latus rectums be

\frac{32}{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:

Answer: (D)

Solution

For

E_{1}

2 a e_{1} = 4

with

e_{1} = 1/ 3

gives

a = 23

, so

a^{2} = 12

, and

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

Its latus rectum is

\frac{2 b ^{2}}{a} = \frac{16}{2 3} = \frac{8}{3}

.

For

E_{2}

(major axis along

y

, semi-major

B

, semi-minor

A

), the latus rectum is

\frac{2 A ^{2}}{B}

, and the product condition gives

\frac{8}{3} \cdot \frac{2 A ^{2}}{B} = \frac{32}{3} ⟹ A^{2} = 2 B .

Since

E_{2}

also has eccentricity

1/ 3

\frac{A ^{2}}{B ^{2}} = \frac{2}{3}

, so

2 B = \frac{2}{3} B^{2} ⟹ B = 3, A^{2} = 6.

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

and

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

.

Solving simultaneously:

x^{2} = 12 - 1.5 y^{2} = 6 - \frac{2}{3} y^{2}

, which gives

y^{2} = \frac{36}{5}

and

x^{2} = \frac{6}{5}

.

The four intersection points

(\pm 6/5, \pm 6/ 5)

form a rectangle with sides

2 6/5

and

12/ 5

, so its area is

4 \cdot \frac{6}{5} \cdot \frac{6}{5} = \frac{24 6}{5} .

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option