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 ?
Select ALL correct options:
A
$S Q_1=2$
B
$Q_1 Q_2=\frac{3 \sqrt{10}}{5}$
C
$P Q_1=3$
D
$S Q_2=1$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
⚠ Partially correct. Some answers are missing.
Solution
Let $P_1\left(t^2, 2 t\right)$ then tangent at $P_1$
$$
t y=x+t^2
$$
Since it passes through ( $-2,1$ )
$$
\begin{array}{r}
\therefore \quad t^2-t-2=0 \\
\quad \therefore \quad t=2,-1
\end{array}
$$
$\therefore \quad P_1(4,4)$ and $P_2(1,-2)$
$$
\therefore \quad S P_1: 4 x-3 y-4=0
$$
and $S P_2: x-1=0$
and for $Q_1: \frac{x_1+2}{4}=\frac{y_1-1}{-3}=\frac{-(-8-3-4)}{25}=\frac{3}{5}$
$$
\therefore \quad x_1=\frac{2}{5}, y_1=\frac{-4}{5}
$$
and $Q_2=(1,1)$
So, $S Q_1=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1$
$$
\begin{aligned}
& Q_1 Q_2=\sqrt{\frac{9}{25}+\frac{81}{25}}=\sqrt{\frac{90}{25}}=\frac{3 \sqrt{10}}{5} \\
& P Q_1=\sqrt{\frac{144}{25}+\frac{81}{25}}=3 \\
& S Q_2=1
\end{aligned}
$$
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