If lhe tangent to the curve `x+y =e^(xy)` be parallel to y-axis, then the point of contact is

Options

A. (1, 0)
B. (0,1)
C. (1,1)
D. None

Answer

To solve the problem, we need to find the point of contact where the tangent to the curve \( x + y = e^{xy} \) is parallel to the y-axis. This means that the slope of the tangent line is undefined, which occurs when \( \frac{dx}{dy} = 0 \).

### Step-by-Step Solution:

1. **Differentiate the given equation**: We start with the equation of the curve:
\[ x + y = e^{xy} \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x + y) = \frac{d}{dx}(e^{xy}) \]
This gives us:
\[ 1 + \frac{dy}{dx} = e^{xy} \left( y + x \frac{dy}{dx} \right) \]

2. **Rearrange the equation**: We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 1 + \frac{dy}{dx} = e^{xy}y + e^{xy}x \frac{dy}{dx} \]
\[ 1 = e^{xy}y + (e^{xy}x - 1) \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{1 - e^{xy}y}{e^{xy}x - 1} \]

3. **Set the condition for the tangent to be vertical**: For the tangent to be parallel to the y-axis, we need \( \frac{dx}{dy} = 0 \). This is equivalent to \( \frac{dy}{dx} \) being undefined, which occurs when the denominator is zero:
\[ e^{xy}x - 1 = 0 \]
Therefore, we have:
\[ e^{xy}x = 1 \]

4. **Solve for \( y \)**: We can express \( y \) in terms of \( x \) using the equation \( e^{xy} = \frac{1}{x} \):
\[ xy = \ln\left(\frac{1}{x}\right) = -\ln x \]
Thus,
\[ y = -\frac{\ln x}{x} \]

5. **Substitute back into the original equation**: We substitute \( y \) back into the original equation \( x + y = e^{xy} \):
\[ x - \frac{\ln x}{x} = e^{- \ln x} = \frac{1}{x} \]
Multiplying through by \( x \) gives:
\[ x^2 - \ln x = 1 \]

6. **Finding the point of contact**: We need to find \( x \) such that \( x^2 - \ln x - 1 = 0 \). This is a transcendental equation and may need numerical methods or graphing to solve. However, we can check specific values.

- For \( x = 1 \):
\[ 1^2 - \ln(1) - 1 = 1 - 0 - 1 = 0 \]
So, \( x = 1 \) is a solution.

- Now, substitute \( x = 1 \) back to find \( y \):
\[ y = -\frac{\ln(1)}{1} = 0 \]
Therefore, the point of contact is \( (1, 0) \).

### Final Answer:
The point of contact where the tangent to the curve is parallel to the y-axis is:
\[ \boxed{(1, 0)} \]