JEE Advanced 2025 · Paper 2 · Mathematics Q4

← JEE Advanced 2025 · Paper 2

MathematicsSingle correct+3 / -1Question 4 of 48

Let $S$ denote the locus of the point of intersection of the pair of lines

$4x - 3y = 12\alpha,\qquad 4\alpha x + 3\alpha y = 12,$

where $\alpha$ varies over the set of non-zero real numbers. Let $T$ be the tangent to $S$ passing through the points $(p, 0)$ and $(0, q)$ , $q > 0$ , and parallel to the line $4x - \dfrac{3}{\sqrt2}y = 0$ .

Then the value of $pq$ is

-6\sqrt2

-3\sqrt2

-9\sqrt2

-12\sqrt2

Answer and worked solution

Official answer: A

Locus. From $4x-3y=12\alpha$ and $4x+3y=\frac{12}{\alpha}$ , multiply to eliminate $\alpha$ :
$16x^2-9y^2=144\implies\frac{x^2}9-\frac{y^2}{16}=1.$

Tangent. The slope is $m=\frac{4\sqrt2}{3}$ . Tangents with this slope are $y=mx\pm\sqrt{9m^2-16}=mx\pm4$ . Since $q>0$ , $T$ is $y=mx+4$ .

Intercepts. $q=4$ and $p=-\frac4m=-\frac{3}{\sqrt2}$ , so $pq=-6\sqrt2$ .

Topics:Hyperbola Tangents and normals to conics Locus

← Previous question Next question →

JEE Advanced 2025 · Paper 2, Mathematics question 4