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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".

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    $\begingroup$ you can be clear whose image is required and about what??? $\endgroup$ – gilly Nov 17 '17 at 10:27
  • $\begingroup$ @gilly Edited the question. Image of the parabola about the tangent $\endgroup$ – Archer Nov 17 '17 at 10:31
  • $\begingroup$ The incorrect graph contradicted intuitive sense of what a reflected image is. When error corrected, intuition checked out ok. $\endgroup$ – Narasimham Nov 18 '17 at 13:25
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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 $t$

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

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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:

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

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  • $\begingroup$ My answer matches the one given in the key.\ $\endgroup$ – Archer Nov 17 '17 at 10:32
  • $\begingroup$ You have changed the graph, and now it is alright. $\endgroup$ – Narasimham Nov 17 '17 at 10:49

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