Mathematics - Application of Derivative Question with Solution | TestHub

$\frac{dy}{dx}=\frac{9t^2-1}{6t}.x=-2,y=2\Rightarrow t=-1,\frac{dy}{dx}=-\frac{4}{3}$ The tangent at P: $y-2=-\frac{4}{3}(x+2)\Rightarrow 4x+3y=-2$ $4\left(1-3t^2\right)+3\left(t-3t^3\right)=-2\Rightarrow (t+1{)}^2(3t-2)=0$ $t=\frac{2}{3}$; slope of tangent at Q is $\frac{dy}{dx}=\frac{9\left(\frac{4}{9}\right)-1}{6\left(\frac{2}{3}\right)}=\frac{3}{4}$. Since $-\frac{4}{3}\times \frac{3}{4}=-1$, the tangents are perpendicular.