# Thread: Another question about the parabola...

1. ## Another question about the parabola...

Lol I seem to be stuck with a lot of questions to do with the parabola.

Q: Show that the normals at P (at^2, 2at) and Q (as^2,2as)on the parabola with equation y^2=4ax meet at the point R where the coordinates of R are
[a(t^2+ts+s^2+2)], -ats(t+s)]. This part of the question I've managed to do. It's the next part that I'm stuck on: Given that the line PQ passes through the focus S(a, 0), show that as t and s vary, R lies on the curve with equation y^2=a(x-3a).

I tried to eliminate the variables t and s, but was unsuccessful. Can someone help please?

Btw, can someone let me know what topic this post should come under for future reference?

2. The line PQ has the equation
$\displaystyle \displaystyle\frac{x-at^2}{a(s^2-t^2}=\frac{y-2at}{2a(s-t)}$
or $\displaystyle \displaystyle\frac{x-at^2}{s+t}=\frac{y-2at}{2}$.
If the line passes through the focus, then we have
$\displaystyle \displaystyle\frac{a-at^2}{s+t}=-\frac{2at}{2}\Rightarrow st=-1$.
The coordinates of R are
$\displaystyle x=a(t^2+st+s^2+2)\Rightarrow x=a(t^2+s^2+1)$
$\displaystyle y=-ats(t+s)\Rightarrow y=a(t+s)$.
Now, we have
$\displaystyle y^2=a^2(t^2+s^2+2st)=a^2(t^2+s^2+1-3)=a^2\left(\frac{x}{a}-3\right)=a(x-3a)$