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

Answer 1

The correct option is C $x+4y=8$

Given curve is $y={(1+x)}^{2y}+{\cos }^2\left({\sin }^{-1}x\right)$
At $x=0\to y=1+{\cos }^2\left(0\right)=2$
Thus, the point is $(0,2)$

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

$\Rightarrow y={(1+x)}^{2y}+1-x^2$

Differentiating with respect to ${}^{\prime }{x}^{\prime }$
$y^{\prime }={(1+x)}^{2y}\{\frac{2y}{1+x}+\ln (1+x).2y^{\prime }\}-2x$

${}^{y^{\prime }}{|}_{(0,2)}=4-0$

Equation of normal at $(0,2)is:y-2=-\frac{1}{4}(x-0)$
$\Rightarrow 4y-8=-x$
$\Rightarrow x+4y=8$