JEE Main 2026 Apr 6 Shift 1, Mathematics Q11: Introduction to Hyperbola

JEE Main2026Apr 6, Shift 1Mathematics

If the eccentricity $e$ of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , passing through $(6, 43)$ , satisfies $15 (e^{2} + 1) = 34 e$ , then the length of the latus rectum of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{2 ( a ^{2} + 1 )} = 1$ is:

A
$10$
B
$20$
C
$25$
D
$30$

Solution

From $15 e^{2} - 34 e + 15 = 0$ , factoring gives $(3 e - 5) (5 e - 3) = 0$ , so $e = \frac{5}{3}$ or $\frac{3}{5}$ . Since $e > 1$ for a hyperbola, $e = \frac{5}{3}$ .

Therefore $1 + \frac{b ^{2}}{a ^{2}} = \frac{25}{9} \Rightarrow \frac{b ^{2}}{a ^{2}} = \frac{16}{9}$ .

The point $(6, 43)$ gives $\frac{36}{a ^{2}} - \frac{48}{b ^{2}} = 1$ . Substituting $b^{2} = \frac{16 a ^{2}}{9}$ leads to $a^{2} = 9$ , hence $a = 3$ and $b = 4$ .

For the second hyperbola, latus rectum length $= \frac{2 \cdot 2 ( a ^{2} + 1 )}{b} = \frac{4 ( 9 + 1 )}{4} = 10$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q11: Introduction to Hyperbola

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