Find the curve in which the length of the normal varies as the square of the ordinate.

Answer

To find the curve in which the length of the normal varies as the square of the ordinate, we can follow these steps:

### Step 1: Understand the relationship
The length of the normal to the curve at any point (x, y) is given by the formula:
\[ L = y \cdot \frac{dy}{dx} \]
According to the problem, this length varies as the square of the ordinate (y), which means:
\[ L = k \cdot y^2 \]
where \(k\) is a constant of proportionality.

### Step 2: Set up the equation
From the above relationships, we can equate the two expressions for the length of the normal:
\[ y \cdot \frac{dy}{dx} = k \cdot y^2 \]

### Step 3: Simplify the equation
Assuming \(y \neq 0\), we can divide both sides by \(y\):
\[ \frac{dy}{dx} = k \cdot y \]

### Step 4: Separate variables
We can rearrange the equation to separate the variables:
\[ \frac{dy}{y} = k \cdot dx \]

### Step 5: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dy}{y} = \int k \cdot dx \]
This gives us:
\[ \log |y| = kx + C \]
where \(C\) is the constant of integration.

### Step 6: Exponentiate to solve for y
To solve for \(y\), we exponentiate both sides:
\[ |y| = e^{kx + C} \]
This can be rewritten as:
\[ y = \pm e^{C} e^{kx} \]
Let \(A = \pm e^{C}\), which is a new constant. Thus, we have:
\[ y = A e^{kx} \]

### Final Result
The equation of the curve is:
\[ y = A e^{kx} \]