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JEE 2025 Mathematics PYQ — Let one focus of the hyperbola. H: be at and the corresponding di… | Mathem Solvex | Mathem Solvex

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JEE 2025 — Mathematics PYQ

JEE | Mathematics | 2025

Let one focus of the hyperbola. H: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be at $(10, 0)$ and the corresponding directrix be $x = \frac{9}{10}$ . If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of H, then $9 (e^{2} + l)$ is equal to:

Choose the correct answer:

A.
14
B.
15
C.
16
(Correct Answer)
D.
12

Correct Answer:

Explanation

$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

$Focus = (10, 0)$

$Transverse axis = 2 a$

$Minor axis = 2 b$

$Eccentricity = e$

$Directrix x = \frac{a}{e}$

$a e = 10 and \frac{a}{e} = \frac{9}{10}$

$\Rightarrow a^{2} = 9 and e = \frac{10}{3}$

$Now, (a e)^{2} = a^{2} + b^{2}$

$10 = 9 + b^{2}$

$\Rightarrow b^{2} = 1$

$Latus rectum (l) = \frac{2 b ^{2}}{a} = \frac{2 \times 1}{3}$

$\Rightarrow 9 (e^{2} + l)$

$= 9 (\frac{10}{9} + \frac{2}{3})$

$= 10 + 6$

$= 16$

Explanation

$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

$Focus = (10, 0)$

$Transverse axis = 2 a$

$Minor axis = 2 b$

$Eccentricity = e$

$Directrix x = \frac{a}{e}$

$a e = 10 and \frac{a}{e} = \frac{9}{10}$

$\Rightarrow a^{2} = 9 and e = \frac{10}{3}$

$Now, (a e)^{2} = a^{2} + b^{2}$

$10 = 9 + b^{2}$

$\Rightarrow b^{2} = 1$

$Latus rectum (l) = \frac{2 b ^{2}}{a} = \frac{2 \times 1}{3}$

$\Rightarrow 9 (e^{2} + l)$

$= 9 (\frac{10}{9} + \frac{2}{3})$

$= 10 + 6$

$= 16$

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