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QJEEAdv24Ma6b30: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma6b30 Solution: Consider the ellipse . Let be a point in the first quadrant…

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 \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?

Multiple Choice Options

  • A q=2,p=33q=2, p=3 \sqrt{3}
  • B q=2,p=43q=2, p=4 \sqrt{3}
  • C q=1,p=53q=1, p=5 \sqrt{3}
  • D q=1,p=63q=1, p=6 \sqrt{3}

Step-by-Step Verified Solution

JEE Advanced 2024 Paper 1 Online Mathematics - Ellipse Question 1 English Explanation

Ar⁡(ΔORT)=32\operatorname{Ar}(\Delta \mathrm{ORT})=\frac{3}{2}

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Tanget at (0,2)(0,2) ,

x(0)9+y(2)4=1⇒y=2\frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2 .........(1)

Tangent at (332,−1)\left(\frac{3 \sqrt{3}}{2},-1\right) ,

x(332)9+y(−1)4=1\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 (A)(\mathrm{A}) is Correct.

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