JEE Advanced 2024 Paper 1 Q04 Mathematics Coordinate Geometry Ellipse Hard

JEE Advanced 2024 Paper 1 · Q04 · Ellipse

Consider the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$. Let $S(p, q)$ be a point in the first quadrant such that $\dfrac{p^2}{9} + \dfrac{q^2}{4} > 1$. 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 $\triangle ORT$ is $\dfrac{3}{2}$, then which of the following options is correct?

  1. A. $q = 2,\ p = 3\sqrt{3}$
  2. B. $q = 2,\ p = 4\sqrt{3}$
  3. C. $q = 1,\ p = 5\sqrt{3}$
  4. D. $q = 1,\ p = 6\sqrt{3}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Parameterize $T=(3\cos\theta,2\sin\theta)$ in QIV, so $\sin\theta<0$. $R=(3,0)$, $O=(0,0)$. Area$=\dfrac{1}{2}|3\cdot 2\sin\theta-0|=3|\sin\theta|=\dfrac{3}{2}\Rightarrow|\sin\theta|=\dfrac{1}{2}$, so $\sin\theta=-\dfrac{1}{2}$, $\cos\theta=\dfrac{\sqrt{3}}{2}$, giving $T=\!\left(\dfrac{3\sqrt{3}}{2},-1\right)$. The other tangent point is the minor-axis endpoint $(0,2)$. The chord-of-contact from $S(p,q)$ is $\dfrac{px}{9}+\dfrac{qy}{4}=1$. Plug in $(0,2)$: $\dfrac{q}{2}=1\Rightarrow q=2$. Plug $T$: $\dfrac{p}{9}\!\cdot\!\dfrac{3\sqrt{3}}{2}+\dfrac{2}{4}\!\cdot\!(-1)=1\Rightarrow \dfrac{p\sqrt{3}}{6}=\dfrac{3}{2}\Rightarrow p=3\sqrt{3}$. Option (A).

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