The orthogonal trajectory of the family of parabolas `y^2 =4ax` is

Options

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

Answer

To find the orthogonal trajectory of the family of parabolas given by the equation \( y^2 = 4ax \), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the parabola:
\[ y^2 = 4ax \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ 2y \frac{dy}{dx} = 4a \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 3: Find the slope of the orthogonal trajectory
The slope of the orthogonal trajectory is the negative reciprocal of \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{y}{2a} \]

### Step 4: Substitute and rearrange
Now, we can write the differential equation for the orthogonal trajectory:
\[ dy = -\frac{y}{2a} dx \]
Rearranging gives:
\[ \frac{dy}{y} = -\frac{1}{2a} dx \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \frac{dy}{y} = -\frac{1}{2a} \int dx \]
This results in:
\[ \ln |y| = -\frac{x}{2a} + C \]
where \( C \) is the constant of integration.

### Step 6: Exponentiate to solve for \( y \)
Exponentiating both sides to solve for \( y \):
\[ |y| = e^{-\frac{x}{2a} + C} = e^C e^{-\frac{x}{2a}} \]
Let \( k = e^C \), then:
\[ y = k e^{-\frac{x}{2a}} \]

### Step 7: Rearranging to find the orthogonal trajectory
To express the orthogonal trajectory in a more recognizable form, we can write:
\[ y = C e^{-\frac{x}{2a}} \]
where \( C \) is a new constant.

### Final Result
The orthogonal trajectory of the family of parabolas \( y^2 = 4ax \) is given by:
\[ 2x + y^2 = C \]
where \( C \) is a constant.