Which of the following points lies on the tangent to the curve

Question:

Which of the following points lies on the tangent to the curve $x^{4} e^{y}+2 \sqrt{y+1}=3$ at the point $(1,0)$ ?

1. (1) $(2,2)$

2. (2) $(2,6)$

3. (3) $(-2,6)$

4. (4) $(-2,4)$

Correct Option: , 3

Solution:

The given curve is, $x^{4} \cdot e^{y}+2 \sqrt{y+1}=3$

Differentiating w.r.t. $x$, we get

$\left(4 x^{3}+x^{4} \cdot y^{\prime}\right) e^{y}+\frac{y^{\prime}}{\sqrt{1+y}}=0$

$\Rightarrow\left(\frac{d y}{d x}\right)=\frac{-4 x^{3} e^{y}}{\left(\frac{1}{\sqrt{y+1}}+e^{y} x^{4}\right)}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{(1,0)}=-2$

$\therefore$ Equation of tangent;

$y-0=-2(x-1) \Rightarrow 2 x+y=2$

Only point $(-2,6)$ lies on the tangent.