Coordinate Geometry – Conic Sections – JEE Main 21 Jan 2026 Shift 1

Solution:

For the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$:
$a^2 = 36 \Rightarrow a = 6$
$b^2 = 16 \Rightarrow b = 4$
Eccentricity $e_1$ is given by $b^2 = a^2(1 – e_1^2)$:
$$16 = 36(1 – e_1^2)$$
$$1 – e_1^2 = \frac{16}{36} = \frac{4}{9}$$
$$e_1^2 = 1 – \frac{4}{9} = \frac{5}{9} \Rightarrow e_1 = \frac{\sqrt{5}}{3}$$

Foci of the ellipse are $(\pm ae_1, 0)$:
$$ae_1 = 6 \times \frac{\sqrt{5}}{3} = 2\sqrt{5}$$
Foci are $(\pm 2\sqrt{5}, 0)$.

Let the hyperbola be $\frac{x^2}{A^2} – \frac{y^2}{B^2} = 1$.
Its foci are $(\pm Ae_2, 0)$.
Given eccentricity of hyperbola $e_2 = 5$.
Since foci coincide:
$$Ae_2 = 2\sqrt{5}$$
$$A(5) = 2\sqrt{5} \Rightarrow A = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$$

For hyperbola, relation between $A, B, e_2$ is $B^2 = A^2(e_2^2 – 1)$:
$$B^2 = \left(\frac{2}{\sqrt{5}}\right)^2 (5^2 – 1)$$
$$B^2 = \frac{4}{5} (24) = \frac{96}{5}$$

Length of Latus Rectum ($LR$) is $\frac{2B^2}{A}$:
$$LR = \frac{2(96/5)}{2/\sqrt{5}}$$
$$LR = \frac{192}{5} \times \frac{\sqrt{5}}{2} = \frac{96\sqrt{5}}{5} = \frac{96}{\sqrt{5}}$$

Ans. (3)