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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse

JEE Advanced 2024 · Paper 1 · Mathematics · Ellipse

Consider the ellipse (x^2)/(9)+(y^2)/(4)=1 . Let S(p, q) be a point...

Consider the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1. Let S(p,q)S(p, q) be a point in the first quadrant such that p29+q24>1\frac{p^2}{9}+\frac{q^2}{4}>1. Two tangents are drawn from SS 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 TT in the fourth quadrant. Let RR be the vertex of the ellipse with positive xx-coordinate and OO be the center of the ellipse. If the area of the triangle △ORT\triangle O R T is 32\frac{3}{2}, then which of the following options is correct?Consider the ellipse (x^2)/(9)+(y^2)/(4)=1 . Let S(p, q) be a point in the first quadrant such that

Options

  1. Aq=2,p=33q=2, p=3 \sqrt{3}
  2. Bq=2,p=43q=2, p=4 \sqrt{3}
  3. Cq=1,p=53q=1, p=5 \sqrt{3}
  4. Dq=1,p=63q=1, p=6 \sqrt{3}

Correct answer

A. q=2,p=33q=2, p=3 \sqrt{3}

Step-by-step solution

Consider the ellipse (x^2)/(9)+(y^2)/(4)=1 . Let S(p, q) be a point in the first quadrant such that Ar⁡(ΔORT)=32∣12×3×2sin⁡θ∣=32sin⁡θ=12⇒θ=11π6\operatorname{Ar}(\Delta \mathrm{ORT})=\frac{3}{2} \left|\frac{1}{2} \times 3 \times 2 \sin \theta\right|=\frac{3}{2} \sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{11 \pi}{6} as point T is in third quadrant T(332,−1)\mathrm{T}\left(\frac{3 \sqrt{3}}{2},-1\right) Tanget at (0,2)x(0)9+y(2)4=1⇒y=2...(1)(0,2) \frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2...(1) Tangent at (332,−1)x(332)9+y(−1)4=1...(2)∴\left(\frac{3 \sqrt{3}}{2},-1\right) \frac{x\left(\frac{3 \sqrt{3}}{2}\right)}{9}+\frac{y(-1)}{4}=1...(2) \therefore By solving (1) & (2) ⇒p=33,q=2⇒\Rightarrow \mathrm{p}=3 \sqrt{3}, \mathrm{q}=2 \Rightarrow Option (1) is Correct.
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