Hello Everyone again! I'm trying to do this question and I'm rather stumped on the last part. Parametric Equation is turning into a nightmare for me .

The full question is as follows:

Show that the tangent at the point P, with parameter t, on the curve $\displaystyle x=3t^2$, $\displaystyle y=3t^2$ has equation $\displaystyle y=tx-t^3$. Prove that this tangent will cut the curve again at the point $\displaystyle Q(\frac{3t^2}{4}, - \frac{t^3}{4})$. Find the coordinates of the possible positions of P if the tangent to the curve at P is the normal to the curve at Q.

To prove for the tangent, I solved (is there a better word) the parametric equation into a cartesian equation, which is: $\displaystyle y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$ Implicitly differentiating, the gradient would be $\displaystyle \frac{\sqrt{x}}{\sqrt{3}}=t$ (From solving the first parametric equation for t). Solve then for tangent.

The gradient therefore is correct and shown as $\displaystyle y=tx-t^3$. Intersecting the gradient and the curve, $\displaystyle y=tx-t^3=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$. Solving for $\displaystyle x$, I get $\displaystyle x=3t^2$ or $\displaystyle \frac{3t^2}{4}$ Since the former option is already taken by P, Q will have to take the other option. Using then $\displaystyle x=\frac{3t^2}{4}$, and substituting it into $\displaystyle y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$, I get the value of $\displaystyle y=\pm \frac{t^3}{4}$. Since the positive solution does not satisfy, the negative must be the correct answer.

Since the tangent to the curve at P is the normal to the curve at Q, and the tangent at P is $\displaystyle t$, we can assume that the tangent at Q is $\displaystyle -\frac{1}{t}$. Here I'm stuck, since I'm not sure how to proceed. Can someone help me over this?

Thanks in advance!!