Question

The curve $y+e^{xy}+x=0$ has a tangent parellel to y-axis at a point

• A. $(-1,0)$
• B. $(1,0)$
• C. $(1,1)$
• D. $(0,0)$

Answer 1

Answer: (A)

$(-1,0)$

$\frac{dy}{dx}+e^{xy}[x\frac{dy}{dx}+y]+1=0$
Or
$\frac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$
Or
$\frac{dy}{dx}=\frac{-(1+y.e^{xy})}{1+x.e^{xy}}$
Now
$1+x{e}^{xy}=0$ since the slope of the tangent is ${90}^0$.
Hence
$x{e}^{xy}=-1$
Or
$x(-x-y)=-1$
Or
$x^2+xy=1$
Or
$y=\frac{1-x^2}{x}$
Or
$y=\frac{1}{x}-x$ ...(i)
Considering $y=0$,
$x^2=1$
$x=\pm 1$.
Hence we get 2 points $(1,0)$ and $(-1,0)$.
Out of these 2, only $(-1,0)$ lies on the curve.
Hence the required point is $(-1,0)$.