If the curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16x$ intersect at right angles, then $a^2=$

We have curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16x$,

Let, these two curves intersect at $\left(x_1, y_1\right)$

Slope of tangent at point $\left(x_1, y_1\right)$ in the curve  $\frac{x^2}{a^2}+\frac{y^2}{4}=1$

$\frac{y^2}{4}=1-\frac{x^2}{a^2} \Rightarrow \frac{2y}{4}\times \frac{dy}{dx}=\frac{-2x}{a^2} \Rightarrow {\frac{dy}{dx}}_{\left(x_1, y_1\right)}=m_{1_{\left(x_1, y_1\right)}}=\frac{-4x_1}{y_1{a}^2}$

Slope of tangent at a point in the curve:  $y^3=16x$

$3y^2\frac{dy}{dx}=16 \Rightarrow {\frac{dy}{dx}}_{\left(x_1, y_1\right)}={m_2}_{\left(x_1, y_1\right)}=\frac{16}{3{y_1}^2}$

These curves intersect at right angle

So, $m_1\times m_2=-1 \Rightarrow \frac{-4x_1}{y_1{a}^2}\times \frac{16}{3{y_1}^2}=-1 \Rightarrow \frac{-64x_1}{3a^2{y_1}^3}=-1 \Rightarrow \frac{-64x_1}{3a^2\times 16x_1}=-1$

$\Rightarrow a^2=\frac{4}{3}$