How to find the mirror image of a parabola with respect to a given line? In a question asked yesterday, the OP wanted to know the answer to the following question in JEE Advanced 2015:

Let the curve C be the mirror image of the parabola $y^2 = 4x$ with respect to the line $x + y + 4 = 0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y = -5$, then what is the distance between $A$ and $B$?

More specifically, the OP wanted to know how to find the equation of the mirror image of the given parabola. However, the question was initially unclear and the thread was put on hold. Since the OP is eager to know the answer and I already prepared a detailed explanation, I decided to create a new question, asking the following:

How can we find the equation of the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$?

 A: 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.

A: 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:
$$
x^2 + 8x + 32 = -4y
$$
A: 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,
$$\boxed{x^2+8x+32=-4y\,}$$
A: The homogeneous matrix of the given parabola is $$Q=\begin{bmatrix}0&0&-2\\0&1&0\\-2&0&0\end{bmatrix}.$$ If the mirror line were $x+y=0$, the reflection would send $x\mapsto-y$ and $y\mapsto-x$, so we can build the reflection in $x+y+4=0$ by composing this with a pair of translations. A point on the latter line is $(0,-4)$, so our reflection is $$R=\begin{bmatrix}1&0&0\\0&1&-4\\0&0&1\end{bmatrix}\begin{bmatrix}0&-1&0\\-1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&4\\0&0&1\end{bmatrix}=\begin{bmatrix}0&-1&-4\\-1&0&-4\\0&0&1\end{bmatrix}.$$ The central matrix in the above product is that of reflection in $x+y=0$ and is easily obtained by recalling that the columns of the matrix are images of the basis vectors. From $R$ we can see that this reflection is equivalent to a reflection in the parallel line through the origin followed by a translation that shifts the image of the desired mirror line so that it passes through the origin. Knowing this, you can construct the reflection matrix directly.
The matrix of the transformed parabola is $(R^{-1})^TQ(R^{-1})=R^TQR$ (a reflection is its own inverse), therefore the equation of the reflected parabola is $$\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}1&0&4\\0&0&2\\4&2&32\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=x^2+8x+4y+32=0.$$  
Because a parabola is symmetric, one could instead rotate it about the point of intersection of its axis and the mirror line, which is $(-4,0)$. The construction is similar to that of the reflection matrix, except that the central matrix of the product is a 90° clockwise rotation instead of a reflection. We really want the inverse rotation, however, so we should instead construct a counterclockwise rotation about $(-4,0)$. It is instructive to work through this to see that in the end you get the same equation for the parabola.
