Consider the parabola $y^{2} = 4 x$ . Let S be the focus of the parabola .A pair of tangents drawn to the parabola from the point P = (-2, 1) meet the parabola at $P_{1}$ and $P_{2}$ . Let $Q_{1}$ and $Q_{2}$ be the points on the lines $S P_{1}$ and $S P_{2}$ respectively .such that $P Q_{1}$ is perpendicular to $S P_{1}$ and $P Q_{2}$ is perpendicular to $S P_{2}$ .Then

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$S Q_{1} = 2$

$Q_{1} Q_{2 =} \frac{3 \sqrt{10}}{5}$

$P Q_{1} = 3$

$S Q_{2} = 1$

answer is B, C, D.

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Detailed Solution

Eq.of tangent at ‘t’ : $x - t y + t^{2} = 0$
This passes through (-2,1) $\Rightarrow t^{2} - t - 2 = 0$ , $\Rightarrow t = 2, - 1$
The points are $(t^{2}, 2 t) = (4, 4), (1, - 2)$
$p_{1} = (4, 4) a n d p_{2} = (1, - 2) s = (1, 0) e q . o f s p_{1} : 4 x - 3 y - 4 = 0 e q . o f s p_{2} : x - 1 = 0 Q_{1} = (\frac{2}{5}, \frac{- 4}{5}) a n d Q_{2} = (1, 1) S Q_{1} = 1; S Q_{2} = 1; Q_{1} Q_{2} = \frac{3 \sqrt{2}}{5}; P Q_{1} = 3$