JEE Main 2025 Jan 22 Shift 2, Mathematics Q16: Introduction to Ellipse

JEE Main2025Jan 22, Shift 2Mathematics

Let $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , $a > b$ and $H : \frac{x ^{2}}{A ^{2}} - \frac{y ^{2}}{B ^{2}} = 1$ . Let the distance between the foci of $E$ and the foci of $H$ be $23$ . If $a - A = 2$ , and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$ , then the sum of the lengths of their latus rectums is equal to :

A
10
B
7
C
8
D
9

Solution

For the ellipse $E$ , foci are $(\pm a e, 0)$ so inter-focal distance is $2 a e = 23 \Rightarrow a e = 3$ .

For the hyperbola $H$ , foci are $(\pm A e^{'}, 0)$ so $2 A e^{'} = 23 \Rightarrow A e^{'} = 3$ .

Hence $a e = A e^{'}$ , so $\frac{e}{e ^{'}} = \frac{A}{a} = \frac{1}{3} \Rightarrow a = 3 A$ .

Combined with $a - A = 2$ : $3 A - A = 2 \Rightarrow A = 1$ , $a = 3$ .

From $a e = 3$ : $e = \frac{3}{3} = \frac{1}{3}$ . From $A e^{'} = 3$ : $e^{'} = 3$ .

For the ellipse: $b^{2} = a^{2} (1 - e^{2}) = 9 (1 - \frac{1}{3}) = 6$ .

For the hyperbola: $B^{2} = A^{2} (e^{'2} - 1) = 1 \cdot (3 - 1) = 2$ .

Sum of latus rectum lengths: $\frac{2 b ^{2}}{a} + \frac{2 B ^{2}}{A} = \frac{12}{3} + \frac{4}{1} = 4 + 4 = 8$ .

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JEE Main 2025 Jan 22 Shift 2, Mathematics Q16: Introduction to Ellipse

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