The question is:
"4. Find the equation of the tangent and the equation of the normal to the curve $x = 3t^{2}$, $y = t^{3}$ at the point whose parameter is $t_{1}$. Find the parameter of the point at which the tangent meets the curve again."
By differentiating $x$ and $y$ with respect to $t$ and using the chain rule, we find:
$\dfrac{dy}{dx} = \dfrac{t}{2}$
Using $y - y_{1} = m(x - x_{1})$ we find the equation of the tangent at the point where $t = t_{1}$ to be:
$2y - t_{1}x + t_{1}^{3} = 0$.
For those who are interested, the equation of the normal is:
$t_{1}y + 2x - (t_{1}^{4} + 6t_{1}^{2}) = 0$.
By using the equation of the tangent, and the fact that the line meets the curve at the points $(3t^{2}, t^{3})$, we find:
$2t^{3} - 3t_{1}t^{2} + t_{1}^{3} = 0$
But I am stuck on how to solve this cubic - can anybody help?
Hint: $$ 2t^3-3t_1t^2+t_1^3= (t-t_1)(2t^2-t_1t -t_1^2) $$