Question

Two mutually perpendicular normals are drawn to the parabola $x^2=2 y$ from a point $P$ on $y-$ axis which meets the parabola at $Q$ and $R$. If area of the $\triangle P Q R$, is $S$, then find the value of $8 S$.

Answer 1

$ x^2=2 y $
$2 x=2 y^{\prime} $
$y^{\prime}=x$
$\therefore$ Slope of the normal at $R=\frac{-1}{t_1}=-1$
$\Rightarrow t _1=1 \Rightarrow R \equiv\left(1, \frac{1}{2}\right)$
Now, $\frac{\lambda-\frac{1}{2}}{0-1}=-1 \Rightarrow \lambda=\frac{3}{2}$
$\therefore S =\operatorname{Ar} .(\triangle PQR )=\frac{1}{2} \times 2 \times 1=1 $
$ \Rightarrow 8 S =8$