1. ## The Parabola

Currently stuck on my hw, to be handed in on Monday, wondering if anyone can help?

The normal to the parabola with the equation y^2=4ax at the point P(at^2, 2at) meets the curve again at the point Q (as^2, 2as). Show that s=-t-(2/t).
I'm not sure where the a, x and y have dissappeared to because when I worked out the equation to the normals at P and Q, I couldn't eliminate them.

Can someone shed any light on this please?

2. $\displaystyle y^{2}=4ax$

$\displaystyle P(at^{2}, \;\ 2at), \;\ Q(as^{2}, \;\ 2as)$

The slope of the tangent line is:

$\displaystyle 2yy'=4a$

$\displaystyle y'=\frac{4a}{2y}$

The slope at P is $\displaystyle \frac{4a}{2(2at)}=\frac{1}{t}$

Therefore, the slope of the normal line is -t.

The line equation is $\displaystyle 2at=-t(at^{2})+b$

$\displaystyle y=-tx+2at+at^{3}$

Use the coordinates of Q and solve for s:

$\displaystyle 2as=-t(as^{2})+2at+at^{3}$

$\displaystyle tas^{2}+2as = 2at+at^{3}$

Factor:

$\displaystyle a(s-t)(t^{2}+st+2)$

s=0 when s=t or upon solving the quadratic: $\displaystyle t^{2}+st+2=0$

$\displaystyle \boxed{s = -t-\frac{2}{t}}$