$\frac{{\mathrm{x}}^2}{\mathrm{\alpha }}+\frac{{\mathrm{y}}^2}{4}=1\Rightarrow \frac{2\mathrm{x}}{\mathrm{\alpha }}+\frac{2\mathrm{y}}{4}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-4\mathrm{x}}{\mathrm{\alpha y}}$                                                     ….(i)

${\mathrm{y}}^3=16\mathrm{x}\Rightarrow 3{\mathrm{y}}^2\cdot \frac{\mathrm{dy}}{\mathrm{dx}}=16\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{16}{3{\mathrm{y}}^2}$                ….(ii)

Since curves intersects at right angles

$\therefore \frac{-4\mathrm{x}}{\mathrm{\alpha y}}+\frac{16}{3{\mathrm{y}}^2}=-1\Rightarrow 3{\mathrm{\alpha y}}^3=64\mathrm{x}$

$\Rightarrow \mathrm{\alpha }=\frac{64\mathrm{x}}{3\times 16\mathrm{x}}=\frac{4}{3}$