Mathematics · Ellipse

JEE Advanced 2024 — Paper 1 — Question 4

Consider the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ . Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^{2}}{9}+\frac{q^{2}}{4}>1$ . Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point T in the fourth quadrant. Let R be the vertex of the ellipse with positive $x$ -coordinate and $O$ be the centre of the ellipse. If the area of the triangle $\triangle \mathrm{ORT}$ is $\frac{3}{2}$ , then which of the following options is correct ?

Option A:
$q=2, p=3 \sqrt{3}$
Correct
Option B:
$q=2, p=4 \sqrt{3}$
Option C:
$q=1, p=5 \sqrt{3}$
Option D:
$q=1, p=6 \sqrt{3}$

Answer: A

Step-by-step solution

Let T be $(3 \cos \theta, 2 \sin \theta)$ $\Rightarrow \operatorname{Ar}(\Delta \mathrm{ORT})=\frac{1}{2} \times 3 \times|2 \sin \theta|=\frac{3}{2}$ $\Rightarrow|\sin \theta|=\frac{1}{2}$ $\Rightarrow \quad \mathrm{T}\left(\frac{3 \sqrt{3}}{2},-1\right)$ Also $\mathrm{q}=2 \Rightarrow \mathrm{~S}(\mathrm{p}, 2)$ T: $\frac{\sqrt{3} x}{6}-\frac{y}{4}=1$ $S(p, 2)$ lies on it $\Rightarrow \frac{\sqrt{3} p}{6}-\frac{2}{4}=1$ $p=\frac{3}{2} \times \frac{6}{\sqrt{3}}$ $p=3 \sqrt{3}$ $\Rightarrow p=3 \sqrt{3}, q=2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Ellipse
Topic: Tangents & Normals to ellipse, chord of conatct