JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let the foci of 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:

A
12
B
16
C
$\frac{96}{5}$
D
$245$

Solution

Ellipse: $a = 6, b = 4$ , eccentricity $e_{1} = 1 - 16/36 = 5 /3$ . Foci at $(\pm a e_{1}, 0) = (\pm 25, 0)$ .

For the hyperbola $\frac{x ^{2}}{A ^{2}} - \frac{y ^{2}}{B ^{2}} = 1$ sharing these foci with $e = 5$ : $A \cdot 5 = 25 \Rightarrow A = \frac{2}{5}$ , and $B^{2} = A^{2} (e^{2} - 1) = \frac{4}{5} \cdot 24 = \frac{96}{5}$ .

Latus rectum length $= \frac{2 B ^{2}}{A} = \frac{2 \cdot 96/5}{2/ 5} = \frac{96}{5}$ .

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