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JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Ellipse

Let f(x)=x^2+9, g(x)=(x)/(x-9) and a=f g(10), b=g f(3) . If e and l...

Let

f(x)=x^2+9, g(x)=\frac{x}{x-9}

and

\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)

. If

\mathrm{e}

and

l

denote the eccentricity and the length of the latus rectum of the ellipse

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

, then

8 \mathrm{e}^2+l^2

is equal to.

Options

A8
B16
C6
D12

Correct answer

A. 8

Step-by-step solution

\begin{aligned} & f(x)=x^2+9 \quad g(x)=\frac{x}{x-9} \\ & a=f(g(10))=f\left(\frac{10}{10-9}\right) \\ & =f(10)=109 \\ & b=g(f(3))=g(9+9) \\ & =g(18)=\frac{18}{9}=2 \\ & E: \frac{x^2}{109}+\frac{y^2}{2}=1\end{aligned}

\begin{aligned} & \mathrm{e}^2=1-\frac{2}{109}=\frac{107}{109} \\ & \ell=\frac{2(2)}{\sqrt{109}}=\frac{4}{\sqrt{109}} \\ & 8 \mathrm{e}^2+\ell^2=\frac{8(107)}{109}+\frac{16}{109} \\ & =8\end{aligned}

$

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