JEE Advanced · Mathematics · 14. Ellipse
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 center of the ellipse. If the area of the triangle \(\triangle O R T\) is \(\frac{3}{2}\), then which of the following options is correct?
- A \(q=2, p=3 \sqrt{3}\)
- B \(q=2, p=4 \sqrt{3}\)
- C \(q=1, p=5 \sqrt{3}\)
- D \(q=1, p=6 \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(q=2, p=3 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Ar}(\Delta \mathrm{ORT})=\frac{3}{2}\)
\(\left|\frac{1}{2} \times 3 \times 2 \sin \theta\right|=\frac{3}{2}\)
\(\sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{11 \pi}{6}\) as point T is in third quadrant
\(\mathrm{T}\left(\frac{3 \sqrt{3}}{2},-1\right)\)
Tanget at \((0,2) \frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2...(1)\)
Tangent at \(\left(\frac{3 \sqrt{3}}{2},-1\right) \frac{x\left(\frac{3 \sqrt{3}}{2}\right)}{9}+\frac{y(-1)}{4}=1...(2)\)
\(\therefore\) By solving (1) & (2) \(\Rightarrow \mathrm{p}=3 \sqrt{3}, \mathrm{q}=2\)
\(\Rightarrow\) Option (1) is Correct.
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