The orthogonal trajectories of the system of curves `y^2 =cx^3 are`

Options

A. `9(y+c)^2 =x^3`
B. `y+c = -x^(3/2)/3`
C. `y+c = x^(3/2)/3`
D. none of these

Answer

To find the orthogonal trajectories of the system of curves given by the equation \( y^2 = cx^3 \), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the curves:
\[ y^2 = cx^3 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(cx^3) \]
Using the chain rule on the left side and the power rule on the right side, we get:
\[ 2y \frac{dy}{dx} = 3cx^2 \]

### Step 2: Express \( c \) in terms of \( y \) and \( x \)
From the differentiated equation, we can express \( c \):
\[ c = \frac{2y}{3x^2} \frac{dy}{dx} \]

### Step 3: Substitute \( c \) back into the original equation
Now, substitute this expression for \( c \) back into the original equation \( y^2 = cx^3 \):
\[ y^2 = \left(\frac{2y}{3x^2} \frac{dy}{dx}\right)x^3 \]
This simplifies to:
\[ y^2 = \frac{2y}{3} x \frac{dy}{dx} \]

### Step 4: Rearrange to find the relationship between slopes
We can rearrange this equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{3y^2}{2x} \]

### Step 5: Find the slope of the orthogonal trajectories
For orthogonal trajectories, the product of the slopes of the curves must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Let \( m_1 = \frac{dy}{dx} = \frac{3y^2}{2x} \) and \( m_2 \) be the slope of the orthogonal trajectories. Thus:
\[ \frac{3y^2}{2x} \cdot m_2 = -1 \]
This gives us:
\[ m_2 = -\frac{2x}{3y^2} \]

### Step 6: Set up the differential equation for orthogonal trajectories
We can express this as:
\[ \frac{dy}{dx} = -\frac{2x}{3y^2} \]

### Step 7: Separate variables and integrate
Separating the variables, we have:
\[ 3y^2 dy = -2x dx \]
Integrating both sides:
\[ \int 3y^2 dy = \int -2x dx \]
This results in:
\[ y^3 = -x^2 + C \]

### Step 8: Rearranging the equation
Rearranging gives us the equation of the orthogonal trajectories:
\[ y^3 + x^2 = C \]

### Final Result
Thus, the orthogonal trajectories of the system of curves \( y^2 = cx^3 \) are given by:
\[ y^3 + x^2 = k \]
where \( k \) is a constant.