Deducing the focus of a parabolic mirror

Question

I've been sitting on it for past three hours but I am unable to spot anything interesting. The first part was obvious but I can't answer the second question (b ii). How can I proof this statement k=a?

Answer 1

One way of deducing this from the fact that $PQR$ is isosceles is to use $QR=PR$.

The equation of the tangent at $P(at^2,2at)$ is $ty=x+at^2$. Therefore the $x$ coordinate of $Q$ is $-at^2$, and the length $QR=k+at^2$.

Meanwhile, the distance $PR=\sqrt{(at^2-k)^2+4a^2t^2}$

Setting these equal, we have $$(k+at^2)^2=(at^2-k)^2+4a^2t^2$$ $$\Rightarrow k^2+2kat^2+a^2t^4=a^2t^4-2akt^2+k^2+4a^2t^2$$ $$4akt^2=4a^2t^2\rightarrow k=a$$

Answer 2

You have to write the equation of line $(PR)$. Set $P=(x_0,y_0)$, $\alpha=\alpha_1=\alpha_2$ and observe $\angle\, xRP=2\alpha$.

Now $\tan\alpha$ is the slope of the tangent line to the parabola at $P$. As $\Bigl(\dfrac{y_0}{2a},1\Bigr)$ is a tangent vector to the parabola, we have: $$\tan\alpha=\frac{2a}{y_0},\enspace\text{hence} \enspace \tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{4ay_0}{y_0^2-4a^2}, $$ so that $$\frac{4ay_0}{y_0^2-4a^2}=\frac{0-y_0}{k-x_0},\enspace\text{whence}\enspace k=\frac{4a^2-y_0^2}{4a}+x_0=a-\frac{y_0^2}{4a}+\frac{y_0^2}{4a}=a.$$