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 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, which of the following is/are TRUE?

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

$S Q_{2} = 1$

$S Q_{1} = 2$

$P Q_{1} = 3$

answer is B, C, D.

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Detailed Solution

Let equation of tangent with slope ‘m’ be

$T : y = m x + \frac{1}{m}$

T: Passes through (–2, 1) so $1 = - 2 m + \frac{1}{m} \Rightarrow m = - 1 o r m = \frac{1}{2}$

Points are given by $(\frac{a}{m^{2}}, \frac{2 a}{m})$

So, one point will be (1, –2) and (4, 4)

Let $P_{1} (4, 4)$ and $P_{2} (1, - 2)$

$P_{1} S : 4 x - 3 y - 4 = 0$

$P_{2} S : x - 1 = 0$

$P Q_{1} = | \frac{4 (- 2) - 3 (1) - 4}{5} | = 3$

$S P = \sqrt{10}; P Q_{2} = 3; S Q_{1} = 1 = S Q_{2}$

$\frac{1}{2} (\frac{Q_{1} Q_{2}}{2}) \times \sqrt{10} = \frac{1}{2} \times 3 \times 1$ (Comparing Areas) $\Rightarrow Q_{1} Q_{2} = \frac{2 \times 3}{\sqrt{10}} = \frac{3 \sqrt{10}}{5}$

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