Question

The curve $y - {e^{xy}} + x = 0$ has a vertical tangent at the point?
A.$\left( {1,1} \right)$
B.At no point
C.$\left( {0,1} \right)$
D.None of these

Answer 1

Here, we will first find the first differentiation of the given equation and then simplify it. Since we know that there is a vertical tangent, then the slope of the tangent is ${90^\circ }$. Then we will take the obtained differentiation equals to zero and then replace 0 for $y$ to find the required point.

Complete step-by-step answer:
We are given that the equation is $y - {e^{xy}} + x = 0$.
Differentiating the above equation, we get

$$\Rightarrow \dfrac{{dy}}{{dx}} - {e^{xy}}\left( {x\dfrac{{dy}}{{dx}} + y} \right) + 1 = 0 \\\ \Rightarrow \dfrac{{dy}}{{dx}} - {e^{xy}}x\dfrac{{dy}}{{dx}} + {e^{xy}}y + 1 = 0 \\\ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - x{e^{xy}}} \right) + 1 - y{e^{xy}} = 0 \\\$$

Adding the above equation with $y{e^{xy}} - 1$ on each of the sides, we get

$$\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - x{e^{xy}}} \right) + 1 - y{e^{xy}} - 1 + y{e^{xy}} = y{e^{xy}} - 1 \\\ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - x{e^{xy}}} \right) = y{e^{xy}} - 1 \\\$$

Dividing the above equation by $1 - x{e^{xy}}$ on each of the sides, we get

$$\Rightarrow \dfrac{{dy}}{{dx}}\dfrac{{\left( {1 - x{e^{xy}}} \right)}}{{1 - x{e^{xy}}}} = \dfrac{{y{e^{xy}} - 1}}{{1 - x{e^{xy}}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y{e^{xy}} - 1}}{{1 - x{e^{xy}}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {1 - y{e^{xy}}} \right)}}{{1 - x{e^{xy}}}} \\\$$

Since we know that there is a vertical tangent, then the slope of the tangent is ${90^\circ }$.
So we will have $1 - x{e^{xy}} = 0$.
Adding the above equation with $x{e^{xy}}$ on each side, we get

$$\Rightarrow 1 - x{e^{xy}} + x{e^{xy}} = 0 + x{e^{xy}} \\\ \Rightarrow 1 = x{e^{xy}} \\\ \Rightarrow x{e^{xy}} = 1{\text{ ......eq.}}\left( 1 \right) \\\$$

Taking the point $\left( {1,1} \right)$ for option A in the above equation, we get

$$\Rightarrow 1 \cdot {e^{1 \times 1}} = 1 \\\ \Rightarrow 1 \cdot e = 1 \\\ \Rightarrow e \ne 1 \\\$$

Since we have found out that the point $\left( {1,1} \right)$ does not satisfy the equation $\left( 1 \right)$, option A is incorrect.
We will now take the point $\left( {0,1} \right)$ for option B in the equation $\left( 1 \right)$, we get

$$\Rightarrow 0 \cdot {e^{0 \times 1}} = 1 \\\ \Rightarrow 0 \cdot 1 = 1 \\\ \Rightarrow 0 \ne 1 \\\$$

Thus, the point $\left( {1,1} \right)$ does not satisfy the equation $\left( 1 \right)$, option C is also incorrect.
Replacing 1 for $x$ and 0 for $y$ in the above equation$\left( 1 \right)$, we get

$$\Rightarrow 1 \cdot {e^{1 \times 0}} = 1 \\\ \Rightarrow 1 \cdot 1 = 1 \\\ \Rightarrow 1 = 1 \\\$$

Since ${\text{LHS}} = {\text{RHS}}$, we have the point $\left( {1,0} \right)$, which satisfies the equation $\left( 1 \right)$.
As $\left( {1,0} \right)$ is in none of the options, the option D is correct.

Note: In this question, some students try to solve the obtained equation $x{e^{xy}} = 1$ to solve for $x$ and $y$, which takes a long time. So we should directly put points into the equation and check which option satisfies the equation. We can also solve this question by finding the normal to the curve instead of the tangent. Since we know that the normal to a curve is perpendicular to the tangent at that point. Thus, we will have that the normal at the point will be 0. The normal curve will be calculated by $- \dfrac{{dx}}{{dy}} = \dfrac{{1 - x{e^{xy}}}}{{y{e^{xy}} - 1}}$ and this will be equal to 0 and we will again reach to the equation $x{e^{xy}} = 1$.