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 $(1,0)$ ?
Correct Option: , 2
Solution:
$x^{4} e^{y}+2 \sqrt{y+1}=3$
d.W.r. to $\mathrm{X}$
$x^{4} e^{y} y^{\prime}+e^{y} 4 x^{3}+\frac{2 y^{\prime}}{2 \sqrt{y+1}}=0$
at $\mathrm{P}(1,0)$
$y_{P}^{\prime}+4+y_{P}^{\prime}=0$
$\Rightarrow y_{P}^{\prime}=-2$
Tangent at $\mathrm{P}(1,0)$ is
$y-0=-2(x-1)$
$2 x+y-2$
$(-2,6)$ lies on it