Find the curve for which the length of normal is equal to the radius vector.

Explanation

To find the curve for which the length of the normal is equal to the radius vector, we will follow these steps:

### Step 1: Define the Length of the Normal
The length of the normal to the curve at a point \((x, y)\) is given by:
\[ L_n = y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \]

### Step 2: Define the Radius Vector
The radius vector from the origin to the point \((x, y)\) is given by:
\[ R = \sqrt{x^2 + y^2} \]

### Step 3: Set Up the Equation
According to the problem, the length of the normal is equal to the radius vector:
\[ y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{x^2 + y^2} \]

### Step 4: Square Both Sides
To eliminate the square roots, we square both sides:
\[ y^2 \left(1 + \left(\frac{dy}{dx}\right)^2\right) = x^2 + y^2 \]

### Step 5: Simplify the Equation
Expanding the left-hand side gives:
\[ y^2 + y^2 \left(\frac{dy}{dx}\right)^2 = x^2 + y^2 \]
Subtracting \(y^2\) from both sides results in:
\[ y^2 \left(\frac{dy}{dx}\right)^2 = x^2 \]

### Step 6: Rearranging the Equation
Dividing both sides by \(y^2\) yields:
\[ \left(\frac{dy}{dx}\right)^2 = \frac{x^2}{y^2} \]
Taking the square root gives:
\[ \frac{dy}{dx} = \frac{x}{y} \quad \text{or} \quad \frac{dy}{dx} = -\frac{x}{y} \]

### Step 7: Separate Variables
We will consider both cases separately. Starting with the first case:
\[ \frac{dy}{dx} = \frac{x}{y} \]
Rearranging gives:
\[ y \, dy = x \, dx \]

### Step 8: Integrate Both Sides
Integrating both sides:
\[ \int y \, dy = \int x \, dx \]
This results in:
\[ \frac{y^2}{2} = \frac{x^2}{2} + C \]
Multiplying through by 2 gives:
\[ y^2 - x^2 = 2C \]

### Step 9: Consider the Second Case
Now for the second case:
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Rearranging gives:
\[ y \, dy = -x \, dx \]

### Step 10: Integrate Both Sides Again
Integrating both sides:
\[ \int y \, dy = -\int x \, dx \]
This results in:
\[ \frac{y^2}{2} = -\frac{x^2}{2} + C \]
Multiplying through by 2 gives:
\[ y^2 + x^2 = 2C \]

### Final Result
Thus, we have two equations representing the curves:
1. \(y^2 - x^2 = 2C\)
2. \(x^2 + y^2 = 2C\)