The normal `y=mx-2am-am^(3)` to the parabola `y^(2)=4ax` subtends a right at the vertex if

Options

A. m = 1
B. `m=sqrt2`
C. `m=-sqrt2`
D. `m=1//sqrt2`

Answer

To solve the problem, we need to determine the condition under which the normal to the parabola \(y^2 = 4ax\) subtends a right angle at the vertex.

### Step-by-Step Solution:

1. **Understanding the Parabola**:
The given parabola is \(y^2 = 4ax\). The vertex of this parabola is at the origin (0, 0).

2. **Equation of the Normal**:
The equation of the normal to the parabola at a point \(t_1\) is given by:
\[ y = mx - 2am - am^3 \]
where \(m\) is the slope of the normal.

3. **Condition for Right Angle**:
For the normal to subtend a right angle at the vertex, the slopes of the lines from the vertex to the points where the normal intersects the parabola must satisfy the condition:
\[ m_1 \cdot m_2 = -1 \]
where \(m_1\) and \(m_2\) are the slopes of the lines from the vertex to the points of intersection.

4. **Finding the Points of Intersection**:
The normal intersects the parabola at points corresponding to parameters \(t_1\) and \(t_2\). The coordinates of these points can be expressed as:
\[ (at_1^2, 2at_1) \quad \text{and} \quad (at_2^2, 2at_2) \]

5. **Finding Slopes**:
The slope \(m_1\) from the vertex to the point \((at_1^2, 2at_1)\) is:
\[ m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1} \]
The slope \(m_2\) from the vertex to the point \((at_2^2, 2at_2)\) is:
\[ m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2} \]

6. **Applying the Right Angle Condition**:
The condition for the right angle becomes:
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \]
Simplifying gives:
\[ \frac{4}{t_1 t_2} = -1 \quad \Rightarrow \quad t_1 t_2 = -4 \]

7. **Relating \(t_1\) and \(t_2\)**:
From the normal equation, we can express \(t_2\) in terms of \(t_1\):
\[ t_2 = -t_1 - \frac{2}{t_1} \]

8. **Substituting into the Condition**:
Substitute \(t_2\) into the equation \(t_1 t_2 = -4\):
\[ t_1 \left(-t_1 - \frac{2}{t_1}\right) = -4 \]
This simplifies to:
\[ -t_1^2 - 2 = -4 \quad \Rightarrow \quad t_1^2 = 2 \]

9. **Finding \(t_1\)**:
Thus, we have:
\[ t_1 = \pm \sqrt{2} \]

10. **Finding \(m\)**:
The slope of the normal is given by \(m = -t_1\). Therefore, we have:
\[ m = -\sqrt{2} \quad \text{or} \quad m = \sqrt{2} \]

### Conclusion:
The values of \(m\) for which the normal subtends a right angle at the vertex are:
\[ m = \sqrt{2} \quad \text{and} \quad m = -\sqrt{2} \]