I've been working my way through an old A'Level maths book and am having a lot of difficulty with a problem given in the Chapter on Loci & Parametric Equations:
"Find the equation of the tangent to the curve $ay^2=x^3$ at the point $(at^2,at^3)$ and prove that, apart from one exceptional case, the tangent meets the curve again. Find the coordinates of the the point of intersection. What is the exceptional case."
I am fine up to finding the equation of the tangent: $2y = 3tx - at^3$ and the exception is when $t=0$ (both verified with the answers given at the back of the book). But I can't seem to get to a/the solution for the intersection point which expects the $x$ and $y$ coordinates to be expressed in terms of $a$ & $t$. I quickly get lost in expanding terms... I guess that I am missing a simplifying step.
Any help appreciated!
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{3at^{2}}{2at} = \frac{3t}{2}$$
Equation of tangent: $$y-at^{3}=\frac{3t}{2} (x-at^{2})$$
Let $(as^{2},as^{3})$ be the intersection, then
\begin{align*} as^{3}-at^{3} &= \frac{3t}{2}(as^{2}-at^{2}) \\ 2(s^{3}-t^{3})-3t(s^{2}-t^{2}) &= 0 \\ (s-t)[2(s^{2}+st+t^{2})-3t(s+t)] &= 0 \\ (s-t)(2s^{2}-st-t^{2}) &= 0 \\ (s-t)^{2}(2s+t) &= 0 \end{align*}
The double roots $s=t$ due to tangency, thus another intersection is given by $s=-\frac{t}{2}$,
that is $$\displaystyle \left( \frac{at^{2}}{4},-\frac{at^{3}}{8} \right)$$
$\fbox{except $t=0$}$.