Question

The curve $y-{{e}^{xy}}+x=0$ has the vertical tangent at the point
A. $(1,1)$
B. $(0,1)$
C. $(1,0)$
D. $(0,0)$

Answer 1

Hint: The curve $y-{{e}^{xy}}+x=0$ has the vertical tangent means it is parallel to Y-axis and perpendicular to X-axis. So, from that find the slope of the line and then differentiate the curve $y-{{e}^{xy}}+x=0$. After that, see whether the points satisfy the equation or not. The point which satisfies will be the required answer.

Complete step by step answer:

Here we are given in the problem that the curve has the vertical tangent means is parallel to Y-axis and perpendicular to X-axis.
So since it is at $90{}^\circ $ to X-axis the slope is,
$\dfrac{dy}{dx}=m=\tan 90$
We know that $\tan 90=\infty $.
So $\dfrac{dy}{dx}=m=\infty $
Now taking the curve $y-{{e}^{xy}}+x=0$ and differentiating with respect to $x$ we get,
$\dfrac{dy}{dx}-{{e}^{xy}}(x\dfrac{dy}{dx}+y)+1=0$
Now writing the above in terms of $\dfrac{dy}{dx}$we get,
$\dfrac{dy}{dx}=\dfrac{y{{e}^{xy}}-1}{1-x{{e}^{xy}}}$
Also, we know that $\dfrac{dy}{dx}=\infty $ so $1-x{{e}^{xy}}=0$.
So, $x{{e}^{xy}}=1$
Now we have got an equation let us see weather which point satisfies the equation.
Let us take option (A), here $x=y=1$.
Taking LHS$=(1){{e}^{1}}=e\ne $RHS
Now option (B), here $x=0,y=1$.
LHS$=(0){{e}^{(0)}}=0\ne $RHS
Option (C), $x=1,y=0$
LHS$=(1){{e}^{(0)}}=1=$RHS
Option (D) both are zero which will not satisfy.
Therefore, $(1,0)$ satisfy the equation.
The curve $y-{{e}^{xy}}+x=0$ has the vertical tangent at the point $(1,0)$.
Option C is the correct answer.

Note: Here we have used the core concept which is if the curve has a vertical tangent than it is parallel to Y-axis and perpendicular to X-axis. Also, the point is satisfied. Here we have took $1-x{{e}^{xy}}=0$ because $\dfrac{dy}{dx}=\infty $.
Additional information: The slope gives the measure of steepness and direction of a straight line. If the slopes of two lines on the Cartesian plane are equal, then the lines are parallel to each other. For two lines to be perpendicular the product of their slope must be equal to $-1$. The x and y coordinates of the lines are used to calculate the slope of the lines. It is the ratio of the change in the y-axis to the change in the x-axis.