How to find the mirror image of a parabola with respect to a given line?

First of all, the question in JEE Advanced 2015 can be solved in a number of ways. One could either mirror the parabola, or mirror the line for which $y=-5$ to find two points of intersection. One could use an intuitive approach ("since the slope of the line equals -1, the mirrored parabola has a function description of the form $y = ax^2 + bx + c$"), or use an approach where rotations and translations are applied to arrive at the correct function description of the mirrored parabola. Below I first show one possible approach to find the equation of the mirrored parabola, then I show that the problem can be solved a lot easier when we simply mirror the line for which $y=-5$.

1. Deriving the equation of the mirrored parabola

Consider the point $P(x,y)$, which we want to mirror with respect to $x+y+4=0$. Since the slope of the mirroring line equals $-1$, the slope of any perpendicular line equals $1$. Let us thus define $P'(x+k,y+k)$ the projection of $P$ on the mirroring line, and $P''(x+2k,y+2k)$ the mirrored point. For $P'$ we find:

$$x + k + y + k + 4 = 0 \iff 2k = -x -y -4$$

As such, we derive that $P''(-y-4,-x-4)$. Filling in the equation of the initial parabola, we get:

$$(-x-4)^2 = 4(-y-4) \iff x^2 + 8x +16 = -4y -16 \iff y = -\frac{x^2}{4}-2x-8$$

To find the intersection with the line for which $y=-5$, we must solve:

$$-\frac{x^2}{4}-2x-8 = -5 \iff x^2 + 8x - 12 = 0 \iff x = 4 \pm 2$$

We thus find a distance of $4$ between the two points of intersection.

2. Deriving the equation of the mirrored line

Since the slope of the mirroring line equals $-1$ and the line we need to mirror is a horizontal one, the mirrored line should be vertical and thus have an equation of the form $x = c$. To find $c$, we only need to find the point of intersection between the two lines:

$$x - 5 + 4 = 0 \iff x = 1$$

Now, all we have to do is find the points of intersection with the original parabola:

$$y^2 = 4 \cdot 1 = 4 \iff y = \pm 2$$

Again, we find a distance of $4$ between the two points of intersection.

If the symmetry line were $y=x$ then the symmetry would be as simple as exchanging $x$ and $y$ coordinates. To reduce to such line you can find an isometry sending $x+y+4=0$ into $x=y$. For example $(x,y)\mapsto (x,-4-y)$. Notice that this isometry is also a symmetry, hence its inverse is itself.

So the symmetry map can be obtained by conjugation. First apply the isometry $(x,y)\mapsto(x,-4-y)$, then the symmetry $(x,y)\mapsto(y,x)$ then the inverse of the isometry. You get: $$ (x,y) \mapsto (x,-4-y) \mapsto (-4-y,x) \mapsto (-4-y,-4-x). $$

So the equation $y^2=4x$ becomes $(-4-x)^2=4(-4-y)$, which is: $$ x^2 + 8x + 32 = -4y $$

Let the general point on the graph of $y^2=4x$ be $(x,y)$. Now, the reflection of a point $(x,y)$ with respect to the line $x+y+4=0$ can be found by the formula,

$$ \dfrac{x_2-x_1}{a}=\dfrac{y_2-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}$$ If we substitute the values accordingly and let the reflected point be $(x',y')$, we get, $$x'=-y-4\\y'=-x-4$$ But we know that the reflected equation is also a parabola and should obey the same equation with the reflected $x$ and $y$ i.e. with $x'$ and $y'$. Therefore, $$(y')^2=4x'$$ Plugging in the obtained values of $x'$ and $y'$, $${(-x-4)}^2=4(-y-4)\\$$ which on simplification reduces to, $$\boxed{x^2+8x+32=-4y\,}$$