If the two curves $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{4}=1$ and ${\mathrm{y}}^3=16\mathrm{x}$ intersect orthogonally then ${\mathrm{a}}^2=$

$$\begin{gathered} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{4}=-1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4\mathrm{x}}{{\mathrm{a}}^2\mathrm{y}}={\mathrm{m}}_1 \\ {\mathrm{y}}^3=16\mathrm{x} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{16}{3{\mathrm{y}}^2}={\mathrm{m}}_2 \\ {\mathrm{m}}_1 {\mathrm{m}}_2=-1 \Rightarrow \frac{-64\mathrm{x}}{3{\mathrm{a}}^2{\mathrm{y}}^3}=-1 \Rightarrow 3{\mathrm{a}}^2=4 \\ \Rightarrow {\mathrm{a}}^2=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$