Mathematics - Parabola Question with Solution | TestHub

Any normal will be $\mathrm{y}+\mathrm{tx}=3\mathrm{t}+\frac{3}{2}{\mathrm{t}}^3$, it passes through (2, 8), so $3{\mathrm{t}}^3+2\mathrm{t}-16=0$ Let $\mathrm{f}(\mathrm{t})=3{\mathrm{t}}^3+2\mathrm{t}-16$ $\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{t})=9{\mathrm{t}}^2+2<0,$ So only one normal.