Question

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

Answer 1

Answer: 8

$x^2=2y$
$2x=2y^{\prime }$
$y^{\prime }=x$
$\therefore$ 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, $\frac{\lambda -\frac{1}{2}}{0-1}=-1\Rightarrow \lambda =\frac{3}{2}$
$\therefore S=\mathrm{Ar}.(△PQR)=\frac{1}{2}\times 2\times 1=1$
$\Rightarrow 8S=8$