If the tangent to the curve $x=1-3t^2,y=t-3t^3$ at the point $P(-2,2)$ meets  the curve again at Q, the angle between the tangent  at P and Q is

$\frac{dy}{dx}=\frac{9t^2-1}{6t}$

$x=-2,y=2\to t=-1,\frac{dy}{dx}=-\frac{4}{3}\mathrm{.............}\left(1\right)$

$\begin{array}{l}TheTangentatP,y-2=-\frac{4}{3}(x+2)\Rightarrow 4x+3y=-2\\ 4(1-3t^2)+3(t-3t^3)=-2\\ \Rightarrow {(t+1)}^2(3t-2)=0\\ t=\frac{2}{3},slopeof\tan gentatQis\frac{dy}{dx}=\frac{9\left(\frac{4}{9}\right)-1}{6\left(\frac{2}{3}\right)}=\frac{3}{4}\mathrm{......}\left(2\right)\end{array}$

(1),(2)  The  tangents  are  perpendicular