The equation of the normal to the curve

Solution:

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 $y=e^{2 y \ln (1+x)}+\cos ^{2}\left(\cos ^{-1} \sqrt{1-x^{2}}\right)$

$y=e^{2 y \ln (1+x)}+\left(\sqrt{1-x^{2}}\right)^{2}$

$y=e^{2 y \ln (1+x)}+\left(1-x^{2}\right)$.......(1)

Now differentiate w.r.t. $x$

$y^{\prime}=e^{2 y \ln (1+x)}\left[2 y \cdot\left(\frac{1}{1+x}\right)+\ln (1+x) \cdot 2 y^{\prime}\right]-2 x$

Put $x=0 \& y=2$

$\mathrm{y}^{\prime}=\mathrm{e}^{2 \times 2 / \mathrm{n} 1}\left[2 \times 2\left(\frac{1}{1+0}\right)+\ln (1+0), 2 \mathrm{y}^{\prime}\right]-2 \times 0$

$y^{\prime}=e^{0}[4+0]-0$

$y^{\prime}=4=$ slope of tangent to the curve

so slope of nomal to the curve $=-\frac{1}{4}\left\{\mathrm{~m}_{1} \mathrm{~m}_{2}=-1\right\}$

Hence equation of normal at $(0,2)$ is

$y-2=-\frac{1}{4}(x-0)$

$\Rightarrow 4 y-8=-x$

$\Rightarrow x+4 y=8$