If the subnormal at any point on the curve `y =3 ^(1-k). x ^(k)` is of constant length the k equals to :

Options

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

Answer

To solve the problem, we need to find the value of \( k \) such that the length of the subnormal at any point on the curve \( y = 3^{1-k} x^k \) is constant.

### Step-by-Step Solution:

1. **Write Down the Curve:**
The given curve is:
\[ y = 3^{1-k} x^k \]

2. **Differentiate to Find the Slope of the Tangent:**
We need to differentiate \( y \) with respect to \( x \) to find the slope of the tangent:
\[ \frac{dy}{dx} = 3^{1-k} \cdot \frac{d}{dx}(x^k) = 3^{1-k} \cdot k x^{k-1} \]
Thus, the slope \( m \) at any point on the curve is:
\[ m = 3^{1-k} \cdot k x^{k-1} \]

3. **Find the Coordinates of the Point on the Curve:**
Let \( P(x_1, y_1) \) be a point on the curve. Then:
\[ y_1 = 3^{1-k} x_1^k \]

4. **Calculate the Length of the Subnormal:**
The length of the subnormal \( L \) is given by the formula:
\[ L = y_1 \cdot \left| \frac{dy}{dx} \right|^{-1} \]
Substituting for \( y_1 \) and \( \frac{dy}{dx} \):
\[ L = y_1 \cdot \frac{1}{\left| 3^{1-k} \cdot k x^{k-1} \right|} = \frac{3^{1-k} x_1^k}{3^{1-k} \cdot k x_1^{k-1}} = \frac{x_1}{k} \]

5. **Set the Length of the Subnormal to be Constant:**
For \( L \) to be constant, \( \frac{x_1}{k} \) must be constant. This implies that \( k \) must be such that \( 2k - 1 = 0 \) (since the length should not depend on \( x_1 \)):
\[ 2k - 1 = 0 \implies k = \frac{1}{2} \]

6. **Conclusion:**
The value of \( k \) for which the length of the subnormal is constant is:
\[ k = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( k \) is \( \frac{1}{2} \).