The normals at the ends of the latusrectum of the parabola `y^(2)=4ax" are (a, 2a) and (a, -2a)"`.

Options

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

Explanation

To solve the problem, we need to find the angle between the normals drawn at the ends of the latus rectum of the parabola given by the equation \( y^2 = 4ax \). The ends of the latus rectum are given as the points \( (a, 2a) \) and \( (a, -2a) \).

### Step-by-Step Solution:

1. **Identify the points on the parabola**:
The ends of the latus rectum for the parabola \( y^2 = 4ax \) are given as:
- Point P: \( (a, 2a) \)
- Point Q: \( (a, -2a) \)

2. **Equation of the normal at point P**:
The general equation of the normal to the parabola at a point \( (x_1, y_1) \) is given by:
\[ y - y_1 = -\frac{y_1}{2a}(x - x_1) \]
Substituting \( (x_1, y_1) = (a, 2a) \):
\[ y - 2a = -\frac{2a}{2a}(x - a) \]
Simplifying this, we have:
\[ y - 2a = -1(x - a) \implies y - 2a = -x + a \implies x + y = 3a \]
This is the equation of the normal at point P.

3. **Equation of the normal at point Q**:
Similarly, for point Q \( (a, -2a) \):
\[ y - (-2a) = -\frac{-2a}{2a}(x - a) \]
Simplifying this:
\[ y + 2a = 1(x - a) \implies y + 2a = x - a \implies x - y = 3a \]
This is the equation of the normal at point Q.

4. **Finding the slopes of the normals**:
From the equations of the normals:
- For the normal at P: \( x + y = 3a \) can be rewritten as \( y = -x + 3a \) which has a slope \( m_1 = -1 \).
- For the normal at Q: \( x - y = 3a \) can be rewritten as \( y = x - 3a \) which has a slope \( m_2 = 1 \).

5. **Finding the angle between the normals**:
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_1 = -1 \) and \( m_2 = 1 \):
\[ \tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{0} \right| \]
Since the denominator is zero, this indicates that the lines are perpendicular, hence:
\[ \theta = 90^\circ \quad \text{or} \quad \theta = \frac{\pi}{2} \text{ radians} \]

### Final Answer:
The angle between the normals at the ends of the latus rectum of the parabola \( y^2 = 4ax \) is \( \frac{\pi}{2} \) radians.