QJEEM24Ma7700: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7700 Solution: Let e 1 be the eccentricity of the hyperbola x 2 16 − y 2 9 = 1 …

Question

Verified Solution • 2025–26 Session moderate

QJEEM24Ma7700: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7700 Solution: Let e 1 be the eccentricity of the hyperbola x 2 16 − y 2 9 = 1 …

Let $e_{1}$ be the eccentricity of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ and $e_{2}$ be the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ , which passes through the foci of the hyperbola. If $e_{1} e_{2} = 1$ , then the length of the chord of the ellipse parallel to the $x$ -axis and passing through $(0, 2)$ is :

Multiple Choice Options

A 853
B 35
C 45
D 1053

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring

Accepted Answer

$\begin{aligned} H : \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 e_{1} = \frac{5}{4} \\ ∴ e_{1} e_{2} = 1 \Rightarrow e_{2} = \frac{4}{5} \end{aligned}$

Also, ellipse is passing through $(\pm 5, 0)$

$\begin{aligned} ∴ a = 5 and b = 3 \\ E : \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \end{aligned}$

JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Hyperbola Question 12 English Explanation

End point of chord are $(\pm \frac{5 \sqrt{5}}{3}, 2)$

$∴ L_{PQ} = \frac{10 \sqrt{5}}{3}$

QJEEM24Ma7700: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7700 Solution: Let e 1 be the eccentricity of the hyperbola x 2 16 − y 2 9 = 1 …

Multiple Choice Options

Step-by-Step Verified Solution

Related Learning Resources & Exam Preparation