$$P: y^2 = 4x$$ $$L: x+y+4 = 0$$
Find mirror image of P with respect to L and mirror image of L with respect to P.
I know how to find the mirror image of a curve with respect to a line... Mirror image of P with respect to L is $x^2+4y+8x+32=0.$ How to find the mirror image of L with respect to P? Or for that matter mirror image of any curve with respect to a parabola? I'm trying to find some relations for closed form.
EDIT: Someone suggested that this problem was duplicate as another one: find image of parabola with respect to line. Please note that its not. I've seen the solution for that, and understood it.
A parametric equation for the parabola $y^2 = 4x$ is $P(t) = (t^2, 2t)$
Then the slope of the tangent line to the parabola at the point $P(t)$ is
$$\dfrac{dy}{dx} = \dfrac{2 \, dt}{2t \, dt} = \dfrac 1t$$
Thus the equation of the normal line to the parabola at the point $P(t)$ is
\begin{align} y - 2t &= -t(x - t^2) \\ y-2t &= x -tx + t^3\\ tx + y &= 2t + t^3 \end{align}
This line intersects the line of reflection, $x+y=-4$, at
$$ L(t) = \left(\frac{4 + 2t + t^3}{t-1}, -\frac{6t+t^3}{t-1} \right)$$
Thus, $L'(t)$, the reflection of the point $L(t)$ about the parabola, $P(t)$ is
$$L'(t) = 2P(t) - L(t) = \left(\dfrac{t^3 - 2 t^2 - 2 t - 4}{t - 1}, \dfrac{t^3 + 4 t^2 + 2 t}{t - 1}\right)$$