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 need to find the fixed point through which the line joining points \( P \) and \( Q \) always passes, given the normals to the parabola \( y^2 = 4x \) at variable points \( P(t_1^2, 2t_1) \) and \( Q(t_2^2, 2t_2) \).

### Step-by-Step Solution:

1. **Equation of the Normal at Point P**:
The normal to the parabola \( y^2 = 4x \) at the point \( P(t_1^2, 2t_1) \) can be derived using the slope of the tangent. The slope of the tangent at point \( P \) is given by \( -\frac{t_1}{2} \). Therefore, the equation of the normal at point \( P \) is:
\[ y - 2t_1 = \frac{1}{2}(x - t_1^2) \]
Rearranging gives:
\[ y = \frac{1}{2}x + 2t_1 - \frac{t_1^2}{2} \]

2. **Equation of the Normal at Point Q**:
Similarly, for point \( Q(t_2^2, 2t_2) \), the equation of the normal is:
\[ y - 2t_2 = \frac{1}{2}(x - t_2^2) \]
Rearranging gives:
\[ y = \frac{1}{2}x + 2t_2 - \frac{t_2^2}{2} \]

3. **Finding the Intersection Point R**:
To find the point \( R(t^2, 2t) \) where the normals intersect, we set the two equations equal to each other:
\[ \frac{1}{2}x + 2t_1 - \frac{t_1^2}{2} = \frac{1}{2}x + 2t_2 - \frac{t_2^2}{2} \]
This simplifies to:
\[ 2t_1 - \frac{t_1^2}{2} = 2t_2 - \frac{t_2^2}{2} \]
Rearranging gives:
\[ t_1^2 - t_2^2 + 4(t_1 - t_2) = 0 \]
Factoring yields:
\[ (t_1 - t_2)(t_1 + t_2 + 4) = 0 \]
Thus, either \( t_1 = t_2 \) or \( t_1 + t_2 + 4 = 0 \).

4. **Finding the Equation of Line PQ**:
The slope of line \( PQ \) is:
\[ \text{slope} = \frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_2 + t_1} \]
The equation of line \( PQ \) can be written as:
\[ y - 2t_1 = \frac{2}{t_2 + t_1}(x - t_1^2) \]

5. **Finding Fixed Point (α, β)**:
Since the line \( PQ \) always passes through a fixed point, we substitute \( t_1 + t_2 = -4 \) into the equation of line \( PQ \):
\[ y - 2t_1 = \frac{2}{-4}(x - t_1^2) \]
This leads to the conclusion that the line \( PQ \) passes through the point \( (-2, 0) \).

6. **Calculating |α + β|**:
Here, we have \( \alpha = -2 \) and \( \beta = 0 \). Therefore:
\[ \alpha + \beta = -2 + 0 = -2 \]
Taking the modulus:
\[ |\alpha + \beta| = |-2| = 2 \]

### Final Answer:
The value of \( |\alpha + \beta| \) is \( 2 \).