Question

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

$x+4y=8$

Explanation of the correct option.

Step $1$: Find the intersection point.

Given: Curve$y={(1+x)}^{2y}+{\cos }^2\left({\sin }^{-1}\left(x\right)\right)$

For intersecting point put $x=0$ in equation of curve,

$$\begin{gathered} y={(1+0)}^{2y}+{\cos }^2\left(0\right) \\ \Rightarrow y=1^{2y}+1 \\ \Rightarrow y=2 \end{gathered}$$

Hence the point of intersection is $\left(0,2\right)$

Step $2$: Find the slope.

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

Differentiate with respect to $x$.

$\frac{dy}{dx}=2y{(1+x)}^{2y-1}+2{(1+x)}^{2y}\ln (1+x)\frac{dy}{dx}+2\cos \left({\sin }^{-1}\left(x\right)\right)(-\sin ({\sin }^{-1}\left(x\right)\left)\right)\frac{1}{\sqrt{1-x^2}}$

slope at the intersection $\left(0,2\right)$ point

$\Rightarrow \frac{dy}{dx}=2\left(2\right){\left(1\right)}^1+0$

$\Rightarrow \frac{dy}{dx}=4$

Since the slope of perpendicular is $-\frac{1}{m}$.

Therefore the slope of the required equation is $-\frac{1}{4}$.

Step $3$: Find the equation of the line.

We know that equation of line passing through $\left(x_{1,}{y}_1\right)$ is given by,

$y-y_1=m(x-x_1)$

Therefore, the equation of line passing through $\left(0,2\right)$ is,

$$\begin{gathered} y-2=-\frac{1}{4}\left(x-0\right) \\ \Rightarrow 4y-8=-x \\ \Rightarrow 4y+x=8 \end{gathered}$$

Hence Option $\left(\mathrm{C}\right)$ is the correct answer.