Question

The equation of the normal to the curve $y=(1+x{)}^{2y}+{\cos }^2\left({\sin }^{-1}x\right)$ at $x=0$ is :

• A. $y=4x+2$
• B. $x+4y=8$
• C. $y+4x=2$
• D. $2y+x=4$

Answer 1

Answer: (B)

Given equation of curve
$y=(1+x{)}^{2y}+{\cos }^2\left({\sin }^{-1}x\right)$
at $x=0$
$y=(1+0{)}^{2y}+{\cos }^2\left({\sin }^{-1}0\right)$
$y=1+1$
$y=2$
So we have to find the normal at $(0,2)$
Now $y=e^{2y\ln (1+x)}+{\cos }^2\left({\cos }^{-1}\sqrt{1-x^2}\right)$
$y=e^{2y\ln (1+x)}+{\left(\sqrt{1-x^2}\right)}^2$
$y=e^{2y\ln (1+x)}+\left(1-x^2\right)\dots (1)$
Now differentiate w.r.t. $x$
$y^{\prime }=e^{2y\ln (1+x)}\left[2y\cdot \left(\frac{1}{1+x}\right)+\ln (1+x).2y^{\prime }\right]-2x$
Put $x=0\&y=2$
$y^{\prime }=e^{2\times 2ln1}\left[2\times 2\left(\frac{1}{1+0}\right)+\ln (1+0).2y^{\prime }\right]-2\times 0$
$y^{\prime }=e^0[4+0]-0$
$y^{\prime }=4=$ 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
$y-2=-\frac{1}{4}(x-0)$
$\Rightarrow 4y-8=-x$
$\Rightarrow x+4y=8$