JEE Advanced 2024 — Paper 1 — MATH-I solution

JEE Advanced2024Paper 1MATH-I

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 $△ O R T$ is $\frac{3}{2}$ , then which of the following options is correct?

A
$q = 2, p = 33$
B
$q = 2, p = 43$
C
$q = 1, p = 53$
D
$q = 1, p = 63$

Solution

One tangent touches at the end of the minor axis. The tangent at $(0, 2)$ is the horizontal line $y = 2$ , so $S$ lies on this line, giving $q = 2$ .

Let $T = (α, β)$ with $α > 0$ , $β < 0$ . Since $R = (3, 0)$ and $O = (0, 0)$ , the area of $△ O R T$ uses base $O R = 3$ and height $∣ β ∣$ :

$\frac{1}{2} \cdot 3 \cdot ∣ β ∣ = \frac{3}{2} ⟹ ∣ β ∣ = 1 ⟹ β = - 1$

Substituting in the ellipse equation: $\frac{α ^{2}}{9} + \frac{1}{4} = 1$ , so $α^{2} = \frac{27}{4}$ and $α = \frac{3 3}{2}$ .

The tangent at $T$ is $\frac{α x}{9} + \frac{β y}{4} = 1$ , i.e. $\frac{( 3 3 /2 ) x}{9} - \frac{y}{4} = 1$ . Substituting $S (p, 2)$ :

$\frac{p 3}{6} - \frac{1}{2} = 1 ⟹ \frac{p 3}{6} = \frac{3}{2} ⟹ p = 33$

$∴ p = 33, q = 2$

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