Mirror image of the parabola about a tangent

Question

The mirror image of the parabola $y^2 = 4x$ in the tangent to the parabola at $(1,2)$ $y^2 = 4x$ is?

I have solved this question correctly (by finding the image of the focus and the vertex in the given line and then finding the parabola that possess these parameters) and got the answer to be $(x+1)^2 = 4(y-1)$

In the graph, I have marked all my findings.

I want to know if there's a simpler method to solve this question? But more importantly, what does this reflected parabola signify? It doesn't actually look like a "reflection". For instance, if we find the reflection of any letter with respect to a line mirror, we see that the letter's images are symmetrical about the line but here the parabolas are touching each other and don't look like what we call "reflections". In a nutshell, it contradicts our intuitive sense of "reflection".

Answer 1

Hint:

Find the equation of the tangent at $(1,2)$

Any point on $y^2=4x,$ can be $(t^2,2t)$

Now if the image of $(t^2,2t)$ is $(h,k)$

then $$\left(\dfrac{t^2+h}2,\dfrac{2t+k}2\right)$$ must lie on the tangent.

Eliminate $t$

Answer 2

The mirror image should have a common tangent at $(1,2)$ so the mirror graph is incorrect.

(my earlier comment in error actually for $45^0$ )

After correction of my obvious errors the mirror should have equation like:

$$ 4 (y-1) = (x+1)^2 $$

Answer 3

The equation of the tangent line is $y=x+1$ and we have to reflect $y^2=4x$ across this line.

Shifting origin to $(0,-1)$ we have to reflect $Y^2 = 4(X-1)$ across $Y=X$. As we know the reflection of $(a,b)$ across $Y=X$ results in the point $(b,a)$. Hence the reflected curve is obtained by switching $X$ and $Y$as $X^2=4(Y-1)$ which in the old coordinate system is $(x+1)^2=4(y-1)$