Question

Let $P\left(x_1,y_1\right)$ (where $y_1>0$ ) be any point on the parabola $y^2=4x$, with focus at $S$. Also normal drawn to the parabola at $P$ cuts the circle described on the focal radius of the point $P$ as diameter at $Q$. If length of $PQ$ is $\sqrt{10}$, then find the perpendicular distance of the point $P$ from the line $x+1=0$.

Answer 1

Answer: 0010

$l_{PQ}=l_{SN}$
$l_{PQ}=\sqrt{1+t^2}=\sqrt{10}\Rightarrow t=3\text{ or }-3\text{ (rejected) }$
$P\equiv (9,6)$
Hence, perpendicular distance of the point $P$ from the line $x+1=0$ is $t^2+1=10$.