The orthogonal trajectories of the family of curves `a^(n-1)y = x^n` are given by

Options

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

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
We start with the equation:
\[ a^{n-1} y = x^n \]
Differentiating 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:
\[ a^{n-1} \frac{dy}{dx} = n x^{n-1} \]

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

### Step 3: Find the slope of the orthogonal trajectories
The slope of the orthogonal trajectories is the negative reciprocal of \(\frac{dy}{dx}\):
\[ \frac{dx}{dy} = -\frac{a^{n-1}}{n x^{n-1}} \]

### Step 4: Rearranging the equation
We can write this as:
\[ n x^{n-1} \, dy + a^{n-1} \, dx = 0 \]

### Step 5: Integrate both sides
Now we integrate:
\[ \int n x^{n-1} \, dy = -\int a^{n-1} \, dx \]
This gives us:
\[ n y = -a^{n-1} x + C \]
where \( C \) is the constant of integration.

### Step 6: Rearranging to standard form
Rearranging gives:
\[ n y + a^{n-1} x = C \]
or
\[ n y + x^2 = k \]
where \( k \) is a constant.

### Final Result
Thus, the orthogonal trajectories of the family of curves \( a^{n-1}y = x^n \) are given by:
\[ n y + x^2 = C \]