Find the length of the line 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 line 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 point on the parabola
The vertex of the parabola \( y^2 = 4ax \) is at the origin \( O(0, 0) \). Let the point on the parabola be \( P(x_1, y_1) \).

### Step 2: Relate the coordinates of point P to angle \( \theta \)
The slope of the line segment \( OP \) making an angle \( \theta \) with the x-axis can be expressed as:
\[ \tan(\theta) = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1} \]
From this, we can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = x_1 \tan(\theta) \]

### Step 3: Substitute \( y_1 \) into the parabola equation
The equation of the parabola is:
\[ y^2 = 4ax \]
Substituting \( y_1 = x_1 \tan(\theta) \) into the parabola's equation gives:
\[ (x_1 \tan(\theta))^2 = 4ax_1 \]
This simplifies to:
\[ x_1^2 \tan^2(\theta) = 4ax_1 \]

### Step 4: Solve for \( x_1 \)
Rearranging the equation, we have:
\[ x_1^2 \tan^2(\theta) - 4ax_1 = 0 \]
Factoring out \( x_1 \):
\[ x_1(x_1 \tan^2(\theta) - 4a) = 0 \]
This gives us two solutions:
1. \( x_1 = 0 \) (the vertex itself)
2. \( x_1 = \frac{4a}{\tan^2(\theta)} \)

### Step 5: Find \( y_1 \)
Using the value of \( x_1 \) to find \( y_1 \):
\[ y_1 = x_1 \tan(\theta) = \frac{4a}{\tan^2(\theta)} \tan(\theta) = \frac{4a}{\tan(\theta)} \]

### Step 6: Coordinates of point P
Thus, the coordinates of point \( P \) are:
\[ P\left(\frac{4a}{\tan^2(\theta)}, \frac{4a}{\tan(\theta)}\right) \]

### Step 7: Calculate the length of segment OP
Using the distance formula to find the length of segment \( OP \):
\[ OP = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2} \]
Substituting the coordinates:
\[ OP = \sqrt{\left(\frac{4a}{\tan^2(\theta)}\right)^2 + \left(\frac{4a}{\tan(\theta)}\right)^2} \]
This simplifies to:
\[ OP = \sqrt{\frac{16a^2}{\tan^4(\theta)} + \frac{16a^2}{\tan^2(\theta)}} \]

### Step 8: Factor out common terms
Factoring out \( 16a^2 \):
\[ OP = 4a \sqrt{\frac{16}{\tan^4(\theta)} + \frac{16}{\tan^2(\theta)}} \]
This can be simplified further:
\[ OP = 4a \sqrt{16\left(\frac{1}{\tan^4(\theta)} + \frac{1}{\tan^2(\theta)}\right)} = 4a \cdot 4 \sqrt{\frac{1}{\tan^4(\theta)} + \frac{1}{\tan^2(\theta)}} \]

### Step 9: Final expression
Thus, the length of the line segment \( OP \) is:
\[ OP = \frac{4a \sec^2(\theta)}{\sin^2(\theta)} \]