Q30 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :
- A16
- B$24 \sqrt{5}$
- C12
- Correct answerD$\frac{96}{\sqrt{5}}$
Solution
For ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$: $a = 6$, $b = 4$, $c = \sqrt{36-16} = 2\sqrt{5}$.
Foci of ellipse: $(\pm 2\sqrt{5}, 0)$.
For hyperbola with eccentricity $e = 5$ and foci at $(\pm 2\sqrt{5}, 0)$:
$e = \frac{C}{A} \Rightarrow A = \frac{2\sqrt{5}}{5}$.
$C^2 = A^2 + B^2 \Rightarrow 20 = \frac{4}{5} + B^2 \Rightarrow B^2 = \frac{96}{5}$.
Latus rectum $= \frac{2B^2}{A} = \frac{2 \cdot \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{96}{\sqrt{5}}$.
Foci of ellipse: $(\pm 2\sqrt{5}, 0)$.
For hyperbola with eccentricity $e = 5$ and foci at $(\pm 2\sqrt{5}, 0)$:
$e = \frac{C}{A} \Rightarrow A = \frac{2\sqrt{5}}{5}$.
$C^2 = A^2 + B^2 \Rightarrow 20 = \frac{4}{5} + B^2 \Rightarrow B^2 = \frac{96}{5}$.
Latus rectum $= \frac{2B^2}{A} = \frac{2 \cdot \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{96}{\sqrt{5}}$.
Hence, the correct answer is option: 'D'.