JEE Advanced 2025 Paper 2 Q04 Mathematics Coordinate Geometry Hyperbola Hard

JEE Advanced 2025 Paper 2 · Q04 · Hyperbola

Let $S$ denote the locus of the point of intersection of the pair of lines $$4x - 3y = 12\alpha,$$ $$4\alpha x + 3\alpha y = 12,$$ 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 $4x - \frac{3}{\sqrt{2}} y = 0$.

Then the value of $p \cdot q$ is

  1. A. $-6\sqrt{2}$
  2. B. $-3\sqrt{2}$
  3. C. $-9\sqrt{2}$
  4. D. $-12\sqrt{2}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Eliminate $\alpha$: from line 1, $\alpha = (4x - 3y)/12$. Sub into line 2: $((4x-3y)/12)(4x + 3y) = 12$, so $(4x-3y)(4x+3y) = 144$ $\to$ $16x^2 - 9y^2 = 144$ $\to$ $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Slope of given line $4x - \frac{3}{\sqrt{2}} y = 0$ is $m = \frac{4\sqrt{2}}{3}$, so $m^2 = \frac{32}{9}$. Tangent to $\frac{x^2}{9} - \frac{y^2}{16} = 1$ with slope $m$ has equation $y = mx \pm \sqrt{9m^2 - 16} = mx \pm \sqrt{32-16} = mx \pm 4$. Since $(0, q)$ lies on $T$ with $q > 0$, choose $+4$: $y = \frac{4\sqrt{2}}{3} x + 4$. $y = 0$ gives $p = -\frac{3}{\sqrt{2}}$; $q = 4$. $p \cdot q = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$. Answer (A).

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