If normal to parabola `y^(2) - 4ax =0` at `alpha` point intersec the parabola again such that sum of ordinates of these two point is 3, then semi latus rectum of the parabola is given by

Options

A. `0.5 alpha`
B. `alpha`
C. `-1.5 alpha`
D. `3 alpha`

Answer

To solve the problem, we need to find the semi-latus rectum of the parabola given that the normal at a point intersects the parabola again, and the sum of the ordinates of these two points is 3.

### Step-by-Step Solution:

1. **Identify the Parabola**: The given parabola is \( y^2 = 4ax \). This is a standard form of a parabola that opens to the right.

2. **Point on the Parabola**: Let the point on the parabola be \( P(\alpha^2, 2a\alpha) \) where \( \alpha \) is a parameter.

3. **Equation of the Normal**: The equation of the normal to the parabola at point \( P \) is given by:
\[ y - 2a\alpha = -\frac{1}{2a\alpha} (x - \alpha^2) \]
Rearranging gives:
\[ y = -\frac{1}{2a\alpha}x + \left(2a\alpha + \frac{\alpha^2}{2a\alpha}\right) \]

4. **Finding Intersection Points**: To find where this normal intersects the parabola again, we substitute \( y \) from the normal equation into the parabola's equation:
\[ \left(-\frac{1}{2a\alpha}x + \left(2a\alpha + \frac{\alpha^2}{2a\alpha}\right)\right)^2 = 4ax \]

5. **Sum of Ordinates**: Let the points of intersection be \( P(\alpha^2, 2a\alpha) \) and \( Q(\beta^2, 2a\beta) \). According to the problem, the sum of the ordinates of these two points is:
\[ 2a\alpha + 2a\beta = 3 \]
Simplifying gives:
\[ 2a(\alpha + \beta) = 3 \quad \Rightarrow \quad \alpha + \beta = \frac{3}{2a} \]

6. **Finding \( \beta \)**: From the normal's properties, we know:
\[ \beta = -\alpha - \frac{2}{\alpha} \]
Substituting this into the sum of ordinates equation:
\[ \alpha - \alpha - \frac{2}{\alpha} = \frac{3}{2a} \]
This simplifies to:
\[ -\frac{2}{\alpha} = \frac{3}{2a} \quad \Rightarrow \quad 2a \cdot 2 = -3\alpha \quad \Rightarrow \quad 4a = -3\alpha \quad \Rightarrow \quad \alpha = -\frac{4a}{3} \]

7. **Finding Semi-Latus Rectum**: The semi-latus rectum \( l \) of the parabola is given by \( l = 2a \). Thus, substituting \( \alpha \):
\[ l = 2a = -\frac{8a}{3} \]

8. **Final Answer**: Therefore, the semi-latus rectum of the parabola is:
\[ \text{Semi-latus rectum} = -1.5\alpha \]