JEE Advanced 2024 · Paper 1 · Mathematics Q4

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MathematicsSingle correct+3 / -1Question 4 of 51

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 $\Delta ORT$ is $\dfrac{3}{2}$ , then which of the following options is correct?

q = 2,\ p = 3\sqrt{3}

q = 2,\ p = 4\sqrt{3}

q = 1,\ p = 5\sqrt{3}

q = 1,\ p = 6\sqrt{3}

Answer and worked solution

Official answer: A

First tangent. A tangent at an end of the minor axis is $y = \pm 2$ ; since $S$ is in the first quadrant it is $y = 2$ , so $q = 2$ .

Point $T$ . Write $T = (3\cos\theta, 2\sin\theta)$ in the fourth quadrant. With $O = (0,0)$ and $R = (3,0)$ ,
$[ORT] = \tfrac12\cdot 3\cdot|2\sin\theta| = \tfrac32 \implies \sin\theta = -\tfrac12,\ \cos\theta = \tfrac{\sqrt3}{2},$
so $T = \left(\tfrac{3\sqrt3}{2}, -1\right)$ .

Second tangent at $T$ : $\dfrac{x\cdot\frac{3\sqrt3}{2}}{9} + \dfrac{y(-1)}{4} = 1$ , i.e. $\dfrac{\sqrt3\,x}{6} - \dfrac{y}{4} = 1$ . It passes through $S(p, 2)$ : $\dfrac{\sqrt3\,p}{6} = \dfrac32$ , so $p = 3\sqrt3$ .

Hence $q = 2$ , $p = 3\sqrt3$ .

Topics:Tangents and normals to conics Ellipse

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JEE Advanced 2024 · Paper 1, Mathematics question 4