The orthogonal trajectories of the family of curves `y=Cx^(2)`, (C is an arbitrary constant), is

Options

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

Explanation

To find the orthogonal trajectories of the family of curves given by \( y = Cx^2 \), where \( C \) is an arbitrary constant, we will follow these steps:

### Step 1: Differentiate the given family of curves
We start with the equation of the family of curves:
\[ y = Cx^2 \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = 2Cx \]

### Step 2: Eliminate the constant \( C \)
From the equation \( y = Cx^2 \), we can express \( C \) in terms of \( y \) and \( x \):
\[ C = \frac{y}{x^2} \]
Substituting this value of \( C \) into the derivative:
\[ \frac{dy}{dx} = 2\left(\frac{y}{x^2}\right)x = \frac{2y}{x} \]

### Step 3: Find the differential equation for orthogonal trajectories
The slopes of the orthogonal trajectories are negative reciprocals of the slopes of the original curves. Therefore, we have:
\[ \frac{dy}{dx} = -\frac{x}{2y} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2y \, dy = -x \, dx \]

### Step 5: Integrate both sides
Now we integrate both sides:
\[ \int 2y \, dy = \int -x \, dx \]
This results in:
\[ y^2 = -\frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 6: Rearranging the equation
Rearranging gives:
\[ x^2 + 2y^2 = 2C \]
This represents the family of orthogonal trajectories.

### Final Result
The orthogonal trajectories of the family of curves \( y = Cx^2 \) are given by:
\[ x^2 + 2y^2 = k \]
where \( k = 2C \) is a constant.

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