Given curves are 

$4x^2+a^2{y}^2=4a^2$ ...(i)

and $y^2=16x$ ...(ii)

If the curves intersect at $P\left(\alpha ,\beta \right)$, then 

$\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{4}=1$ and ${\beta }^2=16\alpha .$

On differentiating equation (i), we get,
$\frac{2x}{a^2}+\frac{2y}{4}{y}^{\text{'}}=0$
$\Rightarrow y^{\prime }=-\frac{4x}{a^{2}y}$
$\Rightarrow m_1=\frac{-4\alpha }{a^2\beta }$

On differentiating equation (ii), we get,
$2y{y}^{\text{'}}=16$
$\Rightarrow m_2=\frac{8}{\beta }$

For curves to be orthogonal,
$m_1{m}_2=-1$

i.e. $\left(-\frac{4\alpha }{a^2\beta }\right)\left(\frac{8}{\beta }\right)=-1$
$\Rightarrow 32\alpha =a^2{\beta }^2$
$\Rightarrow 2{\beta }^2=a^2{\beta }^2$
$\Rightarrow a^2=2$
$\Rightarrow a=\pm \sqrt{2}$
Hence, two values of $a$