JEE Advanced 2025 — Paper 2 — MATH-I solution

JEE Advanced2025Paper 2MATH-I

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

$4 x - 3 y = 12 α, 4 α x + 3 α y = 12,$

where $α$ 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 $4 x - \frac{3}{2} y = 0$ .

Then the value of $pq$ is

A
$- 62$
B
$- 32$
C
$- 92$
D
$- 122$

Solution

Step 1: Eliminate $α$ to find $S$ .

From the first line $α = \frac{4 x - 3 y}{12}$ . The second rearranges to $4 x + 3 y = \frac{12}{α}$ . Multiplying the two,

$(4 x - 3 y) (4 x + 3 y) = 144 \Rightarrow 16 x^{2} - 9 y^{2} = 144,$

so $S : \frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ (a hyperbola).

Step 2: Slope of $T$ .

The given line has slope $m = \frac{4 2}{3}$ .

Step 3: Tangent to the hyperbola.

For $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ , the tangent is $y = m x \pm 9 m^{2} - 16$ . With $9 m^{2} = 32$ , we get $32 - 16 = 4$ , so $3 y = 42 x \pm 12$ .

Step 4: Use $q > 0$ .

Setting $x = 0$ requires the $+ 12$ branch: $3 y = 42 x + 12$ . At $y = 0$ : $p = - \frac{3}{2}$ . At $x = 0$ : $q = 4$ .

Step 5: Compute the product.

$pq = - \frac{3}{2} \cdot 4 = - \frac{12}{2} = - 62$ .

$∴$ The correct option is (A).

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