The value of n for which the length of the subnormal of the curve `xy^n = a^(n+1)` is constant

Options

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

Answer

To find the value of \( n \) for which the length of the subnormal of the curve \( xy^n = a^{n+1} \) is constant, we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the curve:
\[ xy^n = a^{n+1} \]
To find the length of the subnormal, we need to differentiate this equation with respect to \( x \).

Using implicit differentiation:
\[ \frac{d}{dx}(xy^n) = \frac{d}{dx}(a^{n+1}) \]
Applying the product rule on the left side:
\[ y^n + x \cdot n y^{n-1} \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ n x y^{n-1} \frac{dy}{dx} = -y^n \]
Thus,
\[ \frac{dy}{dx} = -\frac{y^n}{n x y^{n-1}} = -\frac{y}{n x} \]

### Step 2: Find the length of the subnormal
The length of the subnormal \( L \) is given by the formula:
\[ L = y \cdot \frac{dy}{dx} \]
Substituting the expression for \( \frac{dy}{dx} \):
\[ L = y \cdot \left(-\frac{y}{n x}\right) = -\frac{y^2}{n x} \]

### Step 3: Express \( x \) in terms of \( y \)
From the original curve equation \( xy^n = a^{n+1} \), we can express \( x \) as:
\[ x = \frac{a^{n+1}}{y^n} \]
Substituting this into the expression for \( L \):
\[ L = -\frac{y^2}{n \cdot \frac{a^{n+1}}{y^n}} = -\frac{y^{n+2}}{n a^{n+1}} \]

### Step 4: Set the length of the subnormal to be constant
For \( L \) to be constant, the term \( y^{n+2} \) must not depend on \( y \). This implies that:
\[ n + 2 = 0 \]
Solving for \( n \):
\[ n = -2 \]

### Conclusion
Thus, the value of \( n \) for which the length of the subnormal of the curve is constant is:
\[ \boxed{-2} \]