JEE Main 2025 Apr 2 Shift 1, Mathematics Q2: Introduction to Hyperbola

JEE Main2025Apr 2, Shift 1Mathematics

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:

A
14
B
15
C
16
D
12

Solution

From the focus and directrix data:

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

Multiplying gives $a^{2} = 9$ , and dividing gives $e^{2} = \frac{10}{9}$ , so $e = \frac{10}{3}$ .

Using the hyperbola relation $b^{2} = a^{2} (e^{2} - 1)$ , or equivalently $(a e)^{2} = a^{2} + b^{2}$ :

$10 = 9 + b^{2} \Rightarrow b^{2} = 1$

Latus rectum length:

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

Therefore:

$9 (e^{2} + l) = 9 (\frac{10}{9} + \frac{2}{3}) = 10 + 6 = 16$

$∴$ Option (3) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q2: Introduction to Hyperbola

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