JEE Main 2025 Jan 29 Shift 1, Mathematics Q12: Introduction to Ellipse

JEE Main2025Jan 29, Shift 1Mathematics

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 $A B C D$ equals:

A
$66$
B
$\frac{18 6}{5}$
C
$\frac{12 6}{5}$
D
$\frac{24 6}{5}$

Solution

Find $E_{1}$ : $2 a e = 4$ with $e = 1/ 3$ gives $a = 23$ , so $a^{2} = 12.$ Then $b^{2} = a^{2} (1 - e^{2}) = 12 \cdot \frac{2}{3} = 8.$

Find $E_{2}$ : for $E_{2}$ the major axis is along $y$ , so $A^{2} = B^{2} (1 - e^{2}) = \frac{2}{3} B^{2} .$ Latus-rectum lengths: $ℓ_{1} = \frac{2 b ^{2}}{a} = \frac{16}{2 3} = \frac{8}{3}$ and $ℓ_{2} = \frac{2 A ^{2}}{B} .$

The product condition: $\frac{8}{3} \cdot \frac{2 A ^{2}}{B} = \frac{32}{3} \Rightarrow \frac{A ^{2}}{B} = 2.$ With $A^{2} = \frac{2}{3} B^{2}$ , $\frac{2}{3} B = 2$ , so $B = 3$ and $A^{2} = 6.$

Intersection points: from $\frac{x ^{2}}{12} + \frac{y ^{2}}{8} = 1$ and $\frac{x ^{2}}{6} + \frac{y ^{2}}{9} = 1$ , eliminating gives $x^{2} = \frac{6}{5}$ and $y^{2} = \frac{36}{5} .$

So the four intersections are $(\pm 6/5, \pm 6/ 5)$ , forming a rectangle with sides $2 6/5$ and $12/ 5 .$

Area $= 2 6/5 \cdot \frac{12}{5} = \frac{24 6}{5} .$

Concept behind this question

Introduction to EllipseNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 1, Mathematics Q12: Introduction to Ellipse

Concept behind this question

More previous year questions on Introduction to Ellipse

Practice more Mathematics