Let denote the locus of the point of intersection of the pair of lines , where varies over the set of non-zero real numbers. Let be the tangent to passing through the points and and parallel to the line . Then the value of is [2025]

Question

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 $4 x - \frac{3}{\sqrt{2}} y = 0$ . Then the value of $p q$ is [2025]

Smart Achievers · Accepted Answer

(1)

$4 x - 3 y = 12 α . . . (i)$

$4 x + 3 y = \frac{12}{α} . . . (i i)$

$For elimination of α, multiply equations (i) and (ii)$

$\Rightarrow 16 x^{2} - 9 y^{2} = 144$ $\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 (Hyperbola)$

$Since, T is parallel to 4 x - \frac{3 y}{\sqrt{2}} = 0$

$\Rightarrow m = \frac{4}{3 / \sqrt{2}} = \frac{4 \sqrt{2}}{3}$

$So equation of tangent:$

$y = m x + \sqrt{a^{2} m^{2} - b^{2}} as q > 0$

$\Rightarrow y = \frac{4 \sqrt{2}}{3} x + \sqrt{9 \times \frac{16 \times 2}{9} - 16}$

$\Rightarrow y = \frac{4 \sqrt{2} x}{3} + 4$ $\Rightarrow (p, 0) \Rightarrow 0 = \frac{4 \sqrt{2} p}{3} + 4 \Rightarrow p = - \frac{3}{\sqrt{2}}$

$and (0, q) \Rightarrow q = 4;$ $∴$ $p q = \frac{- 3}{\sqrt{2}} \times 4 = - 6 \sqrt{2}$

Solution

1 $- 6 \sqrt{2}$

2 $- 3 \sqrt{2}$

3 $- 9 \sqrt{2}$

4 $- 12 \sqrt{2}$

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Solution

Q. Let S denote the locus of the point of intersection of the pair of lines 4x-3y=12α, 4αx+3αy=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 4x-32y=0. Then the value of pq is [2025]

1 -62

2 -32

3 -92

4 -122

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1 $- 6 \sqrt{2}$

2 $- 3 \sqrt{2}$

3 $- 9 \sqrt{2}$

4 $- 12 \sqrt{2}$