Let $S$ denote the locus of the point of intersection of the pair of lines \[ \begin{gathered} 4 x-3 y=12 \alpha, \\ 4 \alpha x+3 \alpha y=12, \end{gathered} \] 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 $4 x-\frac{3}{\sqrt{2}} y=0$. Then the value of $p q$ is

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Straight Lines

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

Answer: (A)

Multiply the two equations:

(4 x - 3 y) (4 α x + 3 α y) = 144 α^{2}

.

Expanding:

16 α x^{2} - 9 α y^{2} = 144 α^{2}

.

Dividing by

α

(non-zero):

16 x^{2} - 9 y^{2} = 144

, or

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

.

This is a hyperbola. The tangent with slope

m = \frac{4 2}{3}

(from the given line) is:

y = \frac{4 2}{3} x \pm 9 m^{2} - 16 = \frac{4 2}{3} x \pm 32 - 16 = \frac{4 2}{3} x \pm 4

Since

q > 0

, we take the positive sign:

3 y = 42 x + 12

.

Setting

y = 0

x = p = - \frac{3}{2}

. Setting

x = 0

y = q = 4

.

Therefore

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

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option