Given $y^2=4ax, xy=c^2$

For $y^2=4ax$,

$2y\frac{dy}{dx}=4a$

i.e. ${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{2a}{y_1}$

For $xy=c^2$,

$x\frac{dy}{dx}+y=0$

i.e. ${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=-\frac{y_1}{x_1}$

For orthogonal intersection of both curves

$\frac{2a}{y_1}\times \left(-\frac{y_1}{x_1}\right)=-1$

i.e. $x_1=2a$

Also, ${y_1}^2=4a{x}_1$

$\Rightarrow {y_1}^2=8a^2$

and ${\left(x_1{y}_1\right)}^2={\left(c^2\right)}^2$

$\Rightarrow c^4=4a^2\times 8a^2=32a^4$