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

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

### Step 2: Express \( c \) in terms of \( y \) and \( x \)
From the differentiation, we can express \( c \):
\[ c = \frac{1}{kx^{1-k}} \frac{dy}{dx} \]

### Step 3: Substitute \( c \) back into the original equation
Substituting \( c \) back into the original equation \( y = cx^k \):
\[ y = \left(\frac{1}{kx^{1-k}} \frac{dy}{dx}\right)x^k \]
This simplifies to:
\[ y = \frac{1}{k} x \frac{dy}{dx} \]

### Step 4: Find the slope of the orthogonal trajectories
The orthogonal trajectories will have slopes that are negative reciprocals of the slopes of the original curves. Therefore, we have:
\[ \frac{dy}{dx} = -\frac{k}{xy} \]

### Step 5: Separate the variables
Rearranging gives:
\[ y \, dy = -\frac{k}{x} \, dx \]

### Step 6: Integrate both sides
Integrating both sides:
\[ \int y \, dy = -k \int \frac{1}{x} \, dx \]
This results in:
\[ \frac{y^2}{2} = -k \ln |x| + C \]

### Step 7: Rearranging the equation
Multiplying through by 2 gives:
\[ y^2 = -2k \ln |x| + C' \]
where \( C' = 2C \).

### Step 8: Final form of the orthogonal trajectories
We can rewrite this as:
\[ y^2 + 2k \ln |x| = C'' \]
where \( C'' \) is another constant.

### Conclusion
The orthogonal trajectories to the family of curves \( y = cx^k \) are given by:
\[ y^2 + 2k \ln |x| = C \]