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

Correct answer: C. 16 aligned & ae= 10 and a e =(9)/(10) & a^2=9 and e= 10 3 & Now aligned & (ae)^2=a^2+b^2 & 10=9+b^2 b^2=1 & & = 2 b^2 a =(2(1))/(3) & 9(e^2+ ) & =9((10)/(9)+(2)/(3)) & =10+6 & =16 aligned aligned

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Hyperbola

Let one focus of the hyperbola H: x^2 a^2 - y^2 ~b^2 =1 be at ( 10...

Let one focus of the hyperbola

\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1

be at

(\sqrt{10}, 0) and the corresponding directrix be

\mathrm{x}=\frac{9}{\sqrt{10}}

. If e and

respectively are the eccentricity and the length of the latus rectum of H , then

\left(\mathrm{e}^2+l\right)

is equal to:

Options

Correct answer

C. 16

Step-by-step solution

\begin{aligned} & \mathrm{ae}=\sqrt{10} \text{\text{ and } } \frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{10} \\ & \Rightarrow \mathrm{a}^2=9 \text{ and } \mathrm{e}=\frac{\sqrt{10}}{3} \\ & \text{ Now } \quad \begin{aligned} & (\mathrm{ae})^2=\mathrm{a}^2+\mathrm{b}^2 \\ & 10=9+\mathrm{b}^2 \Rightarrow \mathrm{ b}^2=1 \\ & \\ & \ell=\frac{2 \mathrm{ b}^2}{\mathrm{a}}=\frac{2(1)}{3} \\ \Rightarrow & 9\left(\mathrm{e}^2+\ell\right) \\ & =9\left(\frac{10}{9}+\frac{2}{3}\right) \\ & =10+6 \\ & =16\end{aligned}

\end{aligned}$

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