Mathematics - Parabola Question with Solution | TestHub

Any point on the given parabola is $\left({\mathrm{t}}^2,2\mathrm{t}\right)$.

The equation of the tangent at $(1,2)$ is $\mathrm{x}-\mathrm{y}+1=0$. The image $(h,k)$ of the point $\left(t^2,2t\right)$ in $x-y+1=0$ is

given by $\frac{\mathrm{h}-{\mathrm{t}}^2}{1}=\frac{\mathrm{k}-2\mathrm{t}}{-1}=\frac{-2\left({\mathrm{t}}^2-2\mathrm{t}+1\right)}{1+1}$

$\therefore h=t^2-t^2+2t-1=2t-1$

and $\mathrm{k}=2\mathrm{t}+{\mathrm{t}}^2-2\mathrm{t}+1={\mathrm{t}}^2+1$

Eliminating t from $\mathrm{h}=2\mathrm{t}-1$ and $\mathrm{k}={\mathrm{t}}^2+1$

we get, $(h+1{)}^2=4(k-1)$

The required equation of reflection is $(x+1{)}^2=4(y-1)$