Given equation of curve
$y=(1+x)^{2 y}+\cos ^{2}\left(\sin ^{-1} x\right)$
at $x =0$
$y=(1+0)^{2 y}+\cos ^{2}\left(\sin ^{-1} 0\right)$
$y=1+1$
$y=2$
So we have to find the normal at $(0,2)$
Now $y=e^{2 y \ln (1+x)}+\cos ^{2}\left(\cos ^{-1} \sqrt{1-x^{2}}\right)$
$y=e^{2 y \ln (1+x)}+\left(\sqrt{1-x^{2}}\right)^{2}$
$y = e ^{2 y \ln (1+ x )}+\left(1- x ^{2}\right) \ldots(1)$
Now differentiate w.r.t. $x$
$y'=e^{2 y \ln (1+x)}\left[2 y \cdot\left(\frac{1}{1+x}\right)+\ln (1+x) . 2 y'\right]-2 x$
Put $x=0 \& y=2$
$y'= e ^{2 \times 2 ln 1}\left[2 \times 2\left(\frac{1}{1+0}\right)+\ln (1+0) . 2 y '\right]-2 \times 0$
$y'= e ^{0}[4+0]-0$
$y'=4=$ slope of tangent to the curve
so slope of normal to the curve $=-\frac{1}{4}\left\{ m _{1} m _{2}=-1\right\}$ Hence equation of normal at (0,2) is
$y-2=-\frac{1}{4}(x-0)$
$\Rightarrow 4 y-8=-x$
$\Rightarrow x+4 y=8$