The curves \(\frac{x^2}{a^2}+\frac{y^2}{4}=4\) and \(y^3=16 x\) intersect each other orthogonally, then \(\mathrm{a}^2=\)

\(
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\
& \therefore \frac{1}{a^2} 2 x+\frac{1}{4} 2 y \frac{d y}{d x}=0 \quad \Rightarrow \frac{d y}{d x}=\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)
\end{aligned}
\)
Also \(\mathrm{y}^3=16 \mathrm{x}\)
\(
\therefore 3 y^2 \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}
\)
Since curves intersect orthogonally, from (1) and (2), we write
\(
\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)\left(\frac{16}{3 y^2}\right)=-1
\)
\(\therefore \frac{64 x}{3 a^2 y^3}=1\) and we have \(y^3=16 x\)
\(
\therefore \frac{64 \mathrm{x}}{3 \mathrm{a}^2(16 \mathrm{x})}=1 \Rightarrow \mathrm{a}^2=\frac{4}{3}
\)