Mathematics · Ellipse

JEE Main 2024 — 9 April, Shift 1 — Question 20

Let $f(x)=x^{2}+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\operatorname{fog}(10), \mathrm{b}=\operatorname{gof}(3)$ . If e and 1 denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$ , then $8 e^{2}+l^{2}$ is equal to

Option A:
16
Option B:
8
Correct
Option C:
6
Option D:
12

Answer: B

Step-by-step solution

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

$\mathrm{a}=\mathrm{f}(\mathrm{g}(10))=\mathrm{f}\left(\frac{10}{10-9}\right)$

$=\mathrm{f}(10)=109$

$b=g(f(3))=g(9+9)$

$=\mathrm{g}(18)=\frac{18}{9}=2$

$E: \frac{x^{2}}{109}+\frac{y^{2}}{2}=1$

$\mathrm{e}^{2}=1-\frac{2}{109}=\frac{107}{109}$

$\ell=\frac{2(2)}{\sqrt{109}}=\frac{4}{\sqrt{109}}$

$8 \mathrm{e}^{2}+\ell^{2}=\frac{8(107)}{109}+\frac{16}{109}$ $=8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse