Let normals to the parabola `y^(2)=4x` at variable points `P(t_(1)^(2), 2t_(1)) and Q(t_(2)^(2), 2t_(2))` meet at the point `R(t^(2) 2t)`, then the line joining P and Q always passes through a fixed point `(alpha, beta)`, then the value of `|alpha+beta|` is equal to

Answer

To solve the problem, we will follow these steps:

1. **Identify the points on the parabola**: The points \( P \) and \( Q \) on the parabola \( y^2 = 4x \) are given as \( P(t_1^2, 2t_1) \) and \( Q(t_2^2, 2t_2) \).

2. **Write the equations of the normals**: The equation of the normal to the parabola at point \( P(t_1^2, 2t_1) \) is given by:
\[ t = -t_1 - \frac{2}{t_1} \]
Similarly, for point \( Q(t_2^2, 2t_2) \), the equation of the normal is:
\[ t = -t_2 - \frac{2}{t_2} \]

3. **Set up the equations**: We can express the equations of the normals in a standard form. For point \( P \):
\[ t = -t_1 - \frac{2}{t_1} \implies t_1 t + t_1^2 + 2 = 0 \]
For point \( Q \):
\[ t = -t_2 - \frac{2}{t_2} \implies t_2 t + t_2^2 + 2 = 0 \]

4. **Combine the equations**: Since \( t_1 \) and \( t_2 \) are roots of the equation \( tx + x^2 + 2 = 0 \), we can write the polynomial as:
\[ x^2 + tx + 2 = 0 \]
where \( t \) is the variable.

5. **Find the sum and product of roots**: From the polynomial, we can find:
- Sum of roots \( t_1 + t_2 = -t \)
- Product of roots \( t_1 t_2 = 2 \)

6. **Equation of the line joining points \( P \) and \( Q \)**: The slope of the line joining \( P \) and \( Q \) can be calculated as:
\[ \text{slope} = \frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_2 + t_1} \]
The equation of the line can be expressed as:
\[ y - 2t_1 = \frac{2}{t_2 + t_1}(x - t_1^2) \]

7. **Simplify the line equation**: Rearranging gives us:
\[ (t_2 + t_1)(y - 2t_1) = 2(x - t_1^2) \]
Expanding and rearranging leads to:
\[ (t_2 + t_1)y - 2t_1(t_2 + t_1) - 2x + 2t_1^2 = 0 \]

8. **Substituting the values of \( t_1 \) and \( t_2 \)**: Using \( t_1 + t_2 = -t \) and \( t_1 t_2 = 2 \):
\[ -ty + 2x + 4 = 0 \]
This line passes through the fixed point \( (-2, 0) \).

9. **Identify \( \alpha \) and \( \beta \)**: From the fixed point, we have \( \alpha = -2 \) and \( \beta = 0 \).

10. **Calculate \( | \alpha + \beta | \)**:
\[ | \alpha + \beta | = | -2 + 0 | = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]