Question

Consider the parabola $y^2=4x$. 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 ?

• A. $S{Q}_l=2$
• B. $Q_1{Q}_2=\frac{3\sqrt{10}}{5}$
• C. $P{Q}_1=3$
• D. $S{Q}_2=1$

Answer 1

Answer: (B), (C), (D)

Let equation of tangent with slope ' $m$ ' be

$T:y=mx+\frac{1}{m}$
$T$ : passes through $(-2,1)$ so
$1=-2m+\frac{1}{m}$
$\Rightarrow m=-1\text{ or }m=\frac{1}{2}$
Points are given by $\left(\frac{a}{m^2},\frac{2a}{m}\right)$
So, one point will be $(1,-2)\&(4,4)$
Let $P_1(4,4)\&P_2(1,-2)$
$P_{1}S:4x-3y-4=0$
$P_{2}S:x-1=0$
$P{Q}_1=\left|\frac{4(-2)-3(1)-4}{5}\right|=3$
$SP=\sqrt{10};P{Q}_2=3;S{Q}_1=1=S{Q}_2$
$\frac{1}{2}\left(\frac{Q_1{Q}_2}{2}\right)\times \sqrt{10}=\frac{1}{2}\times 3\times 1\text{ (comparing Areas) }$
$\Rightarrow Q_1{Q}_2=\frac{2\times 3}{\sqrt{10}}=\frac{3\sqrt{10}}{5}$