Find the angle made by a double ordinate of length `8a` at the vertex of the parabola `y^2=4a xdot`

Options

A. `pi//3`
B. `pi//2`
C. `pi//4`
D. `pi//6`

Answer

To solve the problem of finding the angle made by a double ordinate of length \(8a\) at the vertex of the parabola given by the equation \(y^2 = 4ax\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parabola and its Properties**:
The equation of the parabola is \(y^2 = 4ax\). The vertex of this parabola is at the point \((0, 0)\) and the focus is at \((a, 0)\). The length of the latus rectum is \(4a\).

**Hint**: Remember that the vertex of the parabola \(y^2 = 4ax\) is at the origin, and the focus is at \((a, 0)\).

2. **Understanding the Double Ordinate**:
A double ordinate is a line segment perpendicular to the axis of symmetry of the parabola that intersects it at two points. We are given that the length of the double ordinate is \(8a\).

3. **Finding the Coordinates of the Points on the Double Ordinate**:
Let the double ordinate be at \(x = x_0\). Since the length of the double ordinate is \(8a\), the points on the parabola will be at \((x_0, 4a)\) and \((x_0, -4a)\).

4. **Using the Parabola Equation**:
We substitute \(y = 4a\) into the parabola equation to find \(x_0\):
\[ (4a)^2 = 4a \cdot x_0 \]
Simplifying this gives:
\[ 16a^2 = 4ax_0 \implies x_0 = 4a \]

5. **Identifying the Points**:
The coordinates of the points where the double ordinate intersects the parabola are:
\((4a, 4a)\) and \((4a, -4a)\).

6. **Finding the Angle at the Vertex**:
We need to find the angle \(\theta\) made by the lines connecting the vertex \((0, 0)\) to the points \((4a, 4a)\) and \((4a, -4a)\).

7. **Using Trigonometry**:
In triangle \(OAB\) (where \(O\) is the origin, \(A\) is \((4a, 4a)\), and \(B\) is \((4a, -4a)\)):
- The length of \(OA\) (from the origin to point A) is \(4a\).
- The length of \(OB\) (from the origin to point B) is also \(4a\).
- The length of \(AB\) (the distance between points A and B) is \(8a\).

8. **Finding \(\tan \theta\)**:
The angle \(\theta\) can be found using the tangent function:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4a}{4a} = 1 \]
Thus, \(\theta = 45^\circ\).

9. **Finding the Total Angle**:
Since the angle at the vertex is \(2\theta\):
\[ 2\theta = 2 \times 45^\circ = 90^\circ \]
In radians, this is \(\frac{\pi}{2}\).

10. **Conclusion**:
The angle made by the double ordinate of length \(8a\) at the vertex of the parabola \(y^2 = 4ax\) is \(90^\circ\) or \(\frac{\pi}{2}\).

### Final Answer:
The angle made by the double ordinate of length \(8a\) at the vertex of the parabola \(y^2 = 4ax\) is \(\frac{\pi}{2}\) radians or \(90^\circ\).