The orthogonal trajectories to the family of curve `y= cx^(K)` are given by :

Options

A. `x^2+cy ^2` = constant
B. `x^2 +ky^2` = constant
C. ` kx^2 +y^2=` costant
D. `x^2-ky^2 ` = constant

Answer

To find the orthogonal trajectories to the family of curves given by \( y = cx^k \), we will follow these steps:

### Step 1: Differentiate the given family of curves
We start with the equation of the family of curves:
\[ y = cx^k \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = c \cdot k \cdot x^{k-1} \]

### Step 2: Express \( c \) in terms of \( y \) and \( x \)
From the original equation, we can express \( c \) as:
\[ c = \frac{y}{x^k} \]
Substituting this expression for \( c \) into the derivative:
\[ \frac{dy}{dx} = \frac{y}{x^k} \cdot k \cdot x^{k-1} = \frac{ky}{x} \]

### Step 3: Find the slope of the orthogonal trajectories
The orthogonal trajectories will have slopes that are negative reciprocals of the slopes of the original curves. Thus, we set:
\[ \frac{dy}{dx} = -\frac{x}{ky} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ k y \, dy + x \, dx = 0 \]

### Step 5: Integrate both sides
Now, we will integrate both sides:
\[ \int k y \, dy = -\int x \, dx \]
Integrating gives:
\[ \frac{k}{2} y^2 = -\frac{1}{2} x^2 + C \]
Multiplying through by 2 to simplify:
\[ k y^2 + x^2 = C \]

### Final Result
Thus, the orthogonal trajectories to the family of curves \( y = cx^k \) are given by:
\[ k y^2 + x^2 = C \]