Q: Let $P$ be a point on Parabola $y^2= 4x$ and $Q$ be a point on the line $L_1: 2x+y+4=0$. If the line $L: x-y+1=0$ is the perpendicular bisector of $PQ$ ; then the coordinates of $P$ can be ?
My approach : The line $PQ$ is of form $x+y=c$ , and let $P(t²,2t)$. Since $P$ lies on the line we get $c$ in terms of $t$. Now since the lines $L$ and $PQ$ intersect at midpoint of $P$ and $Q$, I got the coordinates of midpoint and hence got $Q$ in terms of $t$. Then since $Q$ lies on the line $L_1$ I got an equation in $t$ which gives two values of $t$. And hence I got the $2$ coordinates of where $P$ can be....$(1,-2)$ and $(9,-6)$
As you may see this method is too cumbersome and I feel I am missing the point of the question..So is there a better way to do it ??
Consider L as a line of reflection and find out the equation of $L_2$ (the mirror image of $L_1$). Then find out P (the intersection point of $L_2$ and the parabola).
Hint: $x+2y+3=0$ ($L_2$)