If the normal drawn form the point on the axis of the parabola `y^(2)=8ax` whhose distance from the focus is 8 a , and which is not parallel to either axes . Makes an angle `theta` with the axis of x, then `theta` is equal to

Options

A. `pi//6`
B. `pi//4`
C. `pi//3`
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the angle \( \theta \) that the normal makes with the x-axis.

### Step-by-Step Solution:

1. **Identify the Parabola and its Focus**:
The given parabola is \( y^2 = 8ax \). The focus of this parabola is at the point \( (a, 0) \).

2. **Determine the Point on the Axis**:
We know that the distance from the focus to the point on the axis is \( 8a \). The point on the axis can be represented as \( (h, 0) \). The distance from the focus \( (a, 0) \) to the point \( (h, 0) \) is given by:
\[ |h - a| = 8a \]
This gives us two possible cases:
- Case 1: \( h - a = 8a \) → \( h = 9a \)
- Case 2: \( h - a = -8a \) → \( h = -7a \)

3. **Equation of the Normal**:
The equation of the normal to the parabola at point \( (h, 0) \) can be expressed as:
\[ y = mx - 2am - am^3 \]
where \( m \) is the slope of the normal.

4. **Substituting the Point**:
Since the normal passes through the point \( (h, 0) \), we substitute \( y = 0 \) and \( x = h \) into the equation:
\[ 0 = mh - 2am - am^3 \]
Rearranging gives:
\[ mh = 2am + am^3 \]

5. **Finding the Slope**:
From the rearranged equation, we can express \( m \):
\[ m(h - 2a) = am^3 \]
This implies:
\[ m^3 - \frac{h - 2a}{a}m = 0 \]
Factoring out \( m \):
\[ m(m^2 - \frac{h - 2a}{a}) = 0 \]
Thus, we have:
\[ m = 0 \quad \text{or} \quad m^2 = \frac{h - 2a}{a} \]

6. **Calculating for Each Case of \( h \)**:
- For \( h = 9a \):
\[ m^2 = \frac{9a - 2a}{a} = \frac{7a}{a} = 7 \quad \Rightarrow \quad m = \sqrt{7} \text{ or } m = -\sqrt{7} \]
- For \( h = -7a \):
\[ m^2 = \frac{-7a - 2a}{a} = \frac{-9a}{a} = -9 \quad \text{(not valid)} \]

7. **Finding the Angle \( \theta \)**:
The angle \( \theta \) that the normal makes with the x-axis is given by:
\[ \tan(\theta) = m \]
Therefore, for \( m = \sqrt{7} \):
\[ \theta = \tan^{-1}(\sqrt{7}) \]
For \( m = -\sqrt{7} \):
\[ \theta = \tan^{-1}(-\sqrt{7}) \quad \text{(which gives an angle in the 4th quadrant)} \]

8. **Final Values of \( \theta \)**:
The angles corresponding to \( m = \sqrt{7} \) and \( m = -\sqrt{7} \) can be expressed in terms of \( \pi \):
- \( \theta_1 = \tan^{-1}(\sqrt{7}) \)
- \( \theta_2 = \pi + \tan^{-1}(\sqrt{7}) \)

### Conclusion:
Thus, the angle \( \theta \) can take two values based on the slope of the normal:
\[ \theta = \tan^{-1}(\sqrt{7}) \text{ or } \theta = \pi + \tan^{-1}(\sqrt{7}) \]