The equation of normal to `x+y=x^(y)`, where it intersects X-axis, is given by

Options

A. `x+y=1`
B. `x-y-1=0`
C. `x-y+1=0`
D. None of these

Answer

To find the equation of the normal to the curve given by \( x + y = x^y \) at the point where it intersects the x-axis, we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ x + y = x^y \]

### Step 2: Differentiate the equation
To find the slope of the normal, we first need to differentiate the equation implicitly with respect to \( x \).

Differentiating both sides:
\[ \frac{d}{dx}(x + y) = \frac{d}{dx}(x^y) \]

Using the product rule on the right side:
\[ 1 + \frac{dy}{dx} = y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate \( \frac{dy}{dx} \):
\[ 1 + \frac{dy}{dx} - y \cdot x^{y-1} = x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Rearranging gives us:
\[ \frac{dy}{dx} (1 - x^y \ln(x)) = y \cdot x^{y-1} - 1 \]
Thus,
\[ \frac{dy}{dx} = \frac{y \cdot x^{y-1} - 1}{1 - x^y \ln(x)} \]

### Step 5: Evaluate at the intersection with the x-axis
At the x-axis, \( y = 0 \). Substitute \( y = 0 \) into the equation:
\[ x + 0 = x^0 \implies x = 1 \]

### Step 6: Substitute \( x = 1 \) and \( y = 0 \) into \( \frac{dy}{dx} \)
Substituting \( x = 1 \) and \( y = 0 \):
\[ \frac{dy}{dx} = \frac{0 \cdot 1^{-1} - 1}{1 - 1^0 \ln(1)} = \frac{-1}{1 - 0} = -1 \]

### Step 7: Find the slope of the normal
The slope of the normal is the negative reciprocal of \( \frac{dy}{dx} \):
\[ \text{slope of normal} = -\left(-1\right) = 1 \]

### Step 8: Write the equation of the normal
Using the point-slope form of the line, where the point is \( (1, 0) \):
\[ y - 0 = 1(x - 1) \]
This simplifies to:
\[ y = x - 1 \]

### Step 9: Rearranging to standard form
Rearranging gives us:
\[ x - y - 1 = 0 \]

### Final Answer
The equation of the normal to the curve at the point where it intersects the x-axis is:
\[ x - y - 1 = 0 \]

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