Mathematics · Ellipse

JEE Main 2024 — 5 April, Shift 1 — Question 13

Let the line $2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0, \mathrm{k}>0$ , intersect the x -axis and y -axis at the points A and B , respectively. If the equation of the circle having the line segment $A B$ as a diameter is $x^{2}+y^{2}-3 x-2 y=0$ and the length of the latus rectum of the ellipse $\mathrm{x}^{2}+9 \mathrm{y}^{2}=\mathrm{k}^{2}$ is $\frac{\mathrm{m}}{\mathrm{n}}$ , where m and n are coprime, then $2 \mathrm{~m}+\mathrm{n}$ is equal to

Option A:
10
Option B:
11
Correct
Option C:
13
Option D:
12

Answer: B

Step-by-step solution

Centre of the circle $=\left(\frac{3}{2}, 1\right)$

Equation of diameter $=2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0$

$2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0$

$\Rightarrow \mathrm{k}=6$

Now, Equation of ellipse becomes

$x^{2}+9 y^{2}=36$

$\frac{x^{2}}{6^{2}}+\frac{y^{2}}{2^{2}}=1$

length of $\mathrm{LR}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{2.2^{2}}{6}=\frac{8}{6}=\frac{4}{3}=\frac{\mathrm{m}}{\mathrm{n}}$

$\therefore 2 \mathrm{~m}+\mathrm{n}=2(4)+3=11$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse