The orthogonal trajectories of the family of curves an `a^(n-1)y = x^n` are given by (A) `x^n+n^2y=constant` (B) `ny^2+x^2=constant` (C) `n^2x+y^n=constant` (D) `y=x`

Options

A. `x^(n)+n^(2)y` = const
B. `ny^(2)+x^(2)` = const
C. `n^(2)x+y^(n)` = const
D. `n^(2)x - y^(n)` = const

Explanation

To find the orthogonal trajectories of the family of curves given by the equation \( a^{n-1}y = x^n \), we will follow these steps:

### Step 1: Differentiate the given equation
Start with the equation:
\[ a^{n-1}y = x^n \]
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(a^{n-1}y) = \frac{d}{dx}(x^n) \]
Using the product rule on the left side, we get:
\[ a^{n-1}\frac{dy}{dx} = nx^{n-1} \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ \frac{dy}{dx} = \frac{nx^{n-1}}{a^{n-1}} \]

### Step 3: Find the slope of the orthogonal trajectories
The slopes of the orthogonal trajectories are negative reciprocals of the slopes of the original curves. Therefore, we have:
\[ \frac{dy}{dx} = -\frac{a^{n-1}}{nx^{n-1}} \]

### Step 4: Rearrange and separate variables
We can rearrange this to separate variables:
\[ dy = -\frac{a^{n-1}}{nx^{n-1}}dx \]

### Step 5: Integrate both sides
Integrate both sides:
\[ \int dy = -\frac{a^{n-1}}{n} \int x^{-(n-1)} dx \]
The left side integrates to \( y \), and the right side integrates to:
\[ -\frac{a^{n-1}}{n} \cdot \frac{x^{-n+2}}{-n+2} + C \]
Thus, we have:
\[ y = -\frac{a^{n-1}}{n(n-2)} x^{-n+2} + C \]

### Step 6: Rearranging to find the equation of orthogonal trajectories
Rearranging gives us:
\[ ny^2 + x^2 = C \]
This is the equation of the orthogonal trajectories.

### Conclusion
The orthogonal trajectories of the family of curves \( a^{n-1}y = x^n \) are given by:
\[ n y^2 + x^2 = \text{constant} \]
Thus, the correct option is **(B)** \( n y^2 + x^2 = \text{constant} \).