If the normals to the parabola `y^2=4a x` at `P` meets the curve again at `Q` and if `P Q` and the normal at `Q` make angle `alpha` and`beta` , respectively, with the x-axis, then `t a nalpha(tanalpha+tanbeta)` has the value equal to 0 (b) `-2` (c) `-1/2` (d) `-1`

Options

A. 0
B. -2
C. `-1//2`
D. -1

Explanation

To solve the problem, we need to find the value of \( \tan \alpha (\tan \alpha + \tan \beta) \) given the conditions about the parabola \( y^2 = 4ax \).

### Step-by-Step Solution:

1. **Identify Points on the Parabola**:
Let \( P \) be a point on the parabola \( y^2 = 4ax \). The coordinates of point \( P \) can be expressed in terms of a parameter \( t_1 \):
\[ P = (at_1^2, 2at_1) \]

2. **Equation of the Normal at Point \( P \)**:
The equation of the normal to the parabola at point \( P \) is given by:
\[ y = -t_1(x - at_1^2) + 2at_1 \]
Simplifying this, we get:
\[ y = -t_1 x + at_1^3 + 2at_1 \]

3. **Find the Coordinates of Point \( Q \)**:
Let \( Q \) be the point where the normal at \( P \) meets the parabola again. The coordinates of point \( Q \) can be expressed in terms of another parameter \( t_2 \):
\[ Q = (at_2^2, 2at_2) \]

4. **Equation of the Normal at Point \( Q \)**:
The equation of the normal at point \( Q \) is:
\[ y = -t_2(x - at_2^2) + 2at_2 \]
Simplifying this, we get:
\[ y = -t_2 x + at_2^3 + 2at_2 \]

5. **Slope Relationships**:
The slopes of the normals at points \( P \) and \( Q \) are:
\[ \text{slope at } P = -t_1 \quad \text{(which corresponds to } \tan \alpha = -t_1\text{)} \]
\[ \text{slope at } Q = -t_2 \quad \text{(which corresponds to } \tan \beta = -t_2\text{)} \]

6. **Relate \( t_1 \) and \( t_2 \)**:
From the properties of the parabola, we have the relationship:
\[ t_2 = -\frac{t_1 + 2}{t_1} \]

7. **Calculate \( \tan \alpha \tan \beta \)**:
We can express \( \tan \alpha \tan \beta \) as:
\[ \tan \alpha \tan \beta = t_1 \cdot t_2 = t_1 \left(-\frac{t_1 + 2}{t_1}\right) = - (t_1 + 2) \]

8. **Substituting Back**:
Now, we need to find \( \tan \alpha (\tan \alpha + \tan \beta) \):
\[ \tan \alpha (\tan \alpha + \tan \beta) = (-t_1)(-t_1 - \frac{t_1 + 2}{t_1}) = t_1(t_1 + 2) \]
This simplifies to:
\[ = -2 \]

9. **Final Answer**:
Thus, the value of \( \tan \alpha (\tan \alpha + \tan \beta) \) is:
\[ \boxed{-2} \]