Mathematics - Application of Derivative Question with Solution | TestHub

$x=3t^2,y=2t^3,\frac{dy}{dx}=t$. The tangent at $t:y-2t^3=t\left(x-3t^2\right)$. The normal at $t_1:t_{1}y+x=2t_1^4+3t_1^2$. The two are identical: comparing we get $t_1=\frac{1}{t}$ and $-t^3=2t_1^3+3t_1$. Eliminating $t_1,t^6=2+3t^2\Rightarrow t^2=2$, $t=\pm \sqrt{2}$