JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Parabola

For parabola $P_1$ with vertex $(3,2)$ and focus $(4,4)$, the axis is along the line joining these points. The directrix is perpendicular to the axis and equidistant from the vertex on the opposite side.

Since the axis has direction $(1,2)$, the directrix is of the form $x+2y=k$. The distance from vertex to focus is $\sqrt{(4-3{)}^2+(4-2{)}^2}=\sqrt{5}$.

The distance from vertex $(3,2)$ to directrix $x+2y=k$ must also be $\sqrt{5}$:


This gives $k=2$ or $k=12$. Since the directrix must be on the opposite side of the vertex from the focus, and checking that $x+2y=12$ passes through the focus (it does: $4+8=12$), we reject $k=12$ and accept $k=2$.

So the directrix of $P_1$ is $D_1:x+2y=2$.

The mirror line is $\ell :x+2y=6$.

For reflection across a line, if $D_1$ and $D_2$ are equidistant from $\ell$, then:


This gives $|c-6|=4$, so $c=10$ or $c=2$. Since $c=2$ is $D_1$ itself, we have $c=10$.

The directrix of $P_2$ is $x+2y=10$.