JEE Mathematics 2026 Question with Solution

Given: The ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$ and the hyperbola is confocal with it. The eccentricity of the hyperbola is $5$.

Find: The length of the latus rectum of the hyperbola.

For the ellipse, $a^2 = 36$ and $b^2 = 16$. Hence the focal distance is

$$c^2 = a^2 - b^2 = 36 - 16 = 20$$

so

$$c = \sqrt{20} = 2\sqrt{5}$$

Therefore, the common foci are at $\left(\pm 2\sqrt{5}, 0\right)$.

Let the hyperbola be

$$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$

For a hyperbola, $c = Ae$. Since $e = 5$ and $c = 2\sqrt{5}$,

$$2\sqrt{5} = 5A$$

so

$$A = \frac{2\sqrt{5}}{5}$$

Now use

$$B^2 = A^2(e^2 - 1)$$

First,

$$A^2 = \left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{4}{5}$$

Thus,

$$B^2 = \frac{4}{5}(25 - 1) = \frac{4}{5} \times 24 = \frac{96}{5}$$

The length of the latus rectum of the hyperbola is

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

Substituting,

$$\text{L.R.} = \frac{2 \times (96/5)}{2\sqrt{5}/5} = \frac{192/5}{2\sqrt{5}/5} = \frac{192}{2\sqrt{5}} = \frac{96}{\sqrt{5}}$$

Therefore, the length of the latus rectum is $\frac{96}{\sqrt{5}}$. The correct option is D.

Confocal conics have the same value of $c$.

For the ellipse,

$$c^2 = a^2 - b^2 = 36 - 16 = 20$$

so

$$c = 2\sqrt{5}$$

For the hyperbola, $c = Ae$. With $e = 5$,

$$A = \frac{c}{e} = \frac{2\sqrt{5}}{5}$$

Also, for a hyperbola,

$$c^2 = A^2 + B^2$$

Hence,

$$20 = \frac{4}{5} + B^2$$

which gives

$$B^2 = 20 - \frac{4}{5} = \frac{96}{5}$$

Now the latus rectum length is

$$\frac{2B^2}{A} = \frac{2 \times (96/5)}{2\sqrt{5}/5} = \frac{96}{\sqrt{5}}$$

Therefore, the correct option is D.