If `x^(m)+y^(m)=1` such that `(dy)/(dx)=-(x)/(y)`. then what should be the value of m ?

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we start with the given equation and the derivative condition.

1. **Given Equation**:
\[ x^m + y^m = 1 \]

2. **Differentiating Both Sides**:
We will differentiate both sides of the equation with respect to \(x\). Using implicit differentiation, we apply the chain rule:
\[ \frac{d}{dx}(x^m) + \frac{d}{dx}(y^m) = \frac{d}{dx}(1) \]
This gives us:
\[ mx^{m-1} + my^{m-1} \frac{dy}{dx} = 0 \]

3. **Substituting the Given Derivative**:
We know from the problem statement that:
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Substituting this into our differentiated equation:
\[ mx^{m-1} + my^{m-1} \left(-\frac{x}{y}\right) = 0 \]

4. **Simplifying the Equation**:
Rearranging the equation gives:
\[ mx^{m-1} - mx^{m-1}y^{m-2} = 0 \]
Factoring out \(mx^{m-1}\):
\[ mx^{m-1}(1 - y^{m-2}) = 0 \]

5. **Finding Possible Values for \(m\)**:
For the product to be zero, either \(mx^{m-1} = 0\) or \(1 - y^{m-2} = 0\). Since \(x\) and \(y\) cannot be zero (as they are part of the equation \(x^m + y^m = 1\)), we focus on:
\[ 1 - y^{m-2} = 0 \implies y^{m-2} = 1 \]
This implies \(y = 1\) or \(m - 2 = 0\) (since \(y\) must be a real number).

6. **Solving for \(m\)**:
If \(y = 1\), substituting back into the original equation \(x^m + 1^m = 1\) gives \(x^m = 0\), which is not possible. Therefore, we set:
\[ m - 2 = 0 \implies m = 2 \]

7. **Conclusion**:
The only value that satisfies the conditions is \(m = 2\).

### Final Answer:
The value of \(m\) is \(2\).