Find the length of the segment joining the vertex of the parabola `y^(2) = 4ax` and a point on the parabola, where the line segment makes an angle `theta` to the x-axis

Answer

To find the length of the segment joining the vertex of the parabola \( y^2 = 4ax \) and a point on the parabola where the line segment makes an angle \( \theta \) with the x-axis, we can follow these steps:

### Step 1: Identify the vertex and the point on the parabola
The vertex of the parabola \( y^2 = 4ax \) is at the origin, which is the point \( O(0, 0) \). We need to find a point \( P(x, y) \) on the parabola that makes an angle \( \theta \) with the x-axis.

### Step 2: Write the equation of the line
The slope \( m \) of the line that makes an angle \( \theta \) with the x-axis is given by \( m = \tan(\theta) \). The equation of the line passing through the origin can be expressed as:
\[ y = mx = x \tan(\theta) \]

### Step 3: Substitute the line equation into the parabola equation
Since the point \( P(x, y) \) lies on the parabola, we can substitute \( y = x \tan(\theta) \) into the parabola equation \( y^2 = 4ax \):
\[ (x \tan(\theta))^2 = 4ax \]
This simplifies to:
\[ x^2 \tan^2(\theta) = 4ax \]

### Step 4: Rearranging the equation
We can rearrange this equation to isolate \( x \):
\[ x^2 \tan^2(\theta) - 4ax = 0 \]
Factoring out \( x \):
\[ x(x \tan^2(\theta) - 4a) = 0 \]
This gives us two solutions: \( x = 0 \) (the vertex) or \( x = 4a \cot^2(\theta) \).

### Step 5: Find the corresponding \( y \) coordinate
Using \( x = 4a \cot^2(\theta) \) in the line equation \( y = x \tan(\theta) \):
\[ y = 4a \cot^2(\theta) \tan(\theta) = 4a \cot(\theta) \]
Thus, the coordinates of point \( P \) are:
\[ P\left(4a \cot^2(\theta), 4a \cot(\theta)\right) \]

### Step 6: Calculate the length of segment \( OP \)
The length \( OP \) can be calculated using the distance formula:
\[ OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ OP = \sqrt{(4a \cot^2(\theta) - 0)^2 + (4a \cot(\theta) - 0)^2} \]
This simplifies to:
\[ OP = \sqrt{(4a \cot^2(\theta))^2 + (4a \cot(\theta))^2} \]
\[ = \sqrt{16a^2 \cot^4(\theta) + 16a^2 \cot^2(\theta)} \]
Factoring out \( 16a^2 \cot^2(\theta) \):
\[ = \sqrt{16a^2 \cot^2(\theta)(\cot^2(\theta) + 1)} \]
Using the identity \( 1 + \cot^2(\theta) = \csc^2(\theta) \):
\[ = \sqrt{16a^2 \cot^2(\theta) \csc^2(\theta)} \]
\[ = 4a \cot(\theta) \csc(\theta) \]

### Final Result
Thus, the length of the segment joining the vertex of the parabola and the point on the parabola is:
\[ OP = 4a \cot(\theta) \csc(\theta) \]