The equation to the orthogonal trajectories of the system of parabolas `y=ax^2` is

Options

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

Answer

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

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

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

### Step 3: Replace \( \frac{dy}{dx} \) with \( -\frac{dx}{dy} \)
To find the orthogonal trajectories, we replace \( \frac{dy}{dx} \) with \( -\frac{dx}{dy} \):
\[ -\frac{dx}{dy} = \frac{2y}{x} \]
This can be rearranged to:
\[ dx = -\frac{2y}{x} dy \]

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

### Step 5: Integrate both sides
Now we integrate both sides:
\[ \int -x \, dx = \int 2y \, dy \]
The left side integrates to:
\[ -\frac{x^2}{2} + C_1 \]
And the right side integrates to:
\[ y^2 + C_2 \]
Setting \( C = C_2 - C_1 \), we have:
\[ -\frac{x^2}{2} = y^2 + C \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ y^2 + \frac{x^2}{2} + C = 0 \]
This can be written as:
\[ y^2 + \frac{x^2}{2} = -C \]
Letting \( -C = C' \), we can express the final equation as:
\[ y^2 + \frac{x^2}{2} = C' \]

### Final Answer
Thus, the equation of the orthogonal trajectories of the system of parabolas \( y = Ax^2 \) is:
\[ y^2 + \frac{x^2}{2} = C \]