In the parabola `y^2 = 4ax`, the length of the chord passing through the vertex and inclined to the x-axis at an angle `theta` is

Options

A. `4acostheta//sin^2theta`
B. `4acostheta//cos^2theta`
C. `asec^2theta`
D. `a"cosec"^2theta`

Answer

To find the length of the chord passing through the vertex of the parabola \( y^2 = 4ax \) and inclined to the x-axis at an angle \( \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Vertex and Coordinates**:
The vertex of the parabola \( y^2 = 4ax \) is at the origin \( O(0, 0) \). Let \( M \) be a point on the chord such that it makes an angle \( \theta \) with the x-axis.

2. **Determine Coordinates of Point M**:
If the length of the chord is \( l \), the coordinates of point \( M \) can be expressed in terms of \( l \) as:
\[ M = (l \cos \theta, l \sin \theta) \]

3. **Substitute into the Parabola Equation**:
Since point \( M \) lies on the parabola, it must satisfy the equation \( y^2 = 4ax \). Substituting the coordinates of \( M \) into the parabola equation gives:
\[ (l \sin \theta)^2 = 4a(l \cos \theta) \]

4. **Simplify the Equation**:
Expanding and rearranging the equation, we have:
\[ l^2 \sin^2 \theta = 4al \cos \theta \]

5. **Factor Out l**:
We can factor out \( l \) (assuming \( l \neq 0 \)):
\[ l \sin^2 \theta = 4a \cos \theta \]

6. **Solve for l**:
Now, solving for \( l \) gives:
\[ l = \frac{4a \cos \theta}{\sin^2 \theta} \]

### Final Result:
The length of the chord passing through the vertex and inclined to the x-axis at an angle \( \theta \) is:
\[ l = \frac{4a \cos \theta}{\sin^2 \theta} \]