The curve given by `x+y=e^(x y)` has a tangent parallel to the `y - axis` at the point `(0,1)` (b) `(1,0)` (c)`(1,1)` (d) none of these

Options

A. `(0,1)`
B. `(1,0)`
C. `(1,1)`
D. none of these

Answer

To solve the problem, we need to find the point on the curve \( x + y = e^{xy} \) where the tangent is parallel to the y-axis. A tangent is parallel to the y-axis when its slope \( \frac{dy}{dx} \) is infinite, which occurs when the denominator of the derivative equals zero.

### Step-by-step Solution:

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

2. **Rearranging the equation:**
Rearranging the differentiated equation, we get:
\[ 1 + \frac{dy}{dx} = e^{xy}y + e^{xy}x\frac{dy}{dx} \]
Now, we can isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} - e^{xy}x\frac{dy}{dx} = e^{xy}y - 1 \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(1 - e^{xy}x) = e^{xy}y - 1 \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{e^{xy}y - 1}{1 - e^{xy}x} \]

3. **Setting the denominator to zero:**
For the tangent to be parallel to the y-axis, we need:
\[ 1 - e^{xy}x = 0 \]
This implies:
\[ e^{xy}x = 1 \quad \Rightarrow \quad e^{xy} = \frac{1}{x} \]

4. **Finding points (0,1), (1,0), (1,1):**
We will check each point to see if it satisfies \( e^{xy} = \frac{1}{x} \).

- **For (0,1):**
\[ e^{0 \cdot 1} = e^0 = 1 \quad \text{and} \quad \frac{1}{0} \quad \text{(undefined)} \]
Not valid.

- **For (1,0):**
\[ e^{1 \cdot 0} = e^0 = 1 \quad \text{and} \quad \frac{1}{1} = 1 \]
Valid.

- **For (1,1):**
\[ e^{1 \cdot 1} = e^1 = e \quad \text{and} \quad \frac{1}{1} = 1 \]
Not valid.

5. **Conclusion:**
The only point where the tangent to the curve is parallel to the y-axis is at \( (1,0) \).

### Final Answer:
The point where the tangent is parallel to the y-axis is \( (1,0) \).

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