TITLE: How to find the mirror image of a parabola with respect to a given line?
QUESTION [9 upvotes]: 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 $x+y+4=0$?

REPLY [4 votes]: 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,
$$\boxed{x^2+8x+32=-4y\,}$$