The chord AC of the parabola `y^(2)=4ax` subtends an angle of `90^(@)` at points B and D on the parabola. If points A, B, C and D are represented by `(at_(i)^(2), 2at_(i)), i=1,2,3,4` respectively, then find the value of `|(t_(2)+t_(4))/(t_(1)+t_(3))|`.

Answer

To solve the problem, we need to analyze the given parabola and the points on it. The parabola is given by the equation \( y^2 = 4ax \). The points A, B, C, and D on the parabola are represented as follows:

- \( A(t_1^2, 2at_1) \)
- \( B(t_2^2, 2at_2) \)
- \( C(t_3^2, 2at_3) \)
- \( D(t_4^2, 2at_4) \)

### Step 1: Find the slopes of the lines AB and BC
The slope of line segment AB is given by:
\[ M_{AB} = \frac{2at_2 - 2at_1}{t_2^2 - t_1^2} = \frac{2a(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2a}{t_2 + t_1} \]

The slope of line segment BC is given by:
\[ M_{BC} = \frac{2at_3 - 2at_2}{t_3^2 - t_2^2} = \frac{2a(t_3 - t_2)}{(t_3 - t_2)(t_3 + t_2)} = \frac{2a}{t_3 + t_2} \]

### Step 2: Set up the equation for the angle subtended at point B
Since the angle subtended at point B is \( 90^\circ \), we have:
\[ M_{AB} \cdot M_{BC} = -1 \]
Substituting the slopes:
\[ \left(\frac{2a}{t_2 + t_1}\right) \cdot \left(\frac{2a}{t_3 + t_2}\right) = -1 \]
This simplifies to:
\[ \frac{4a^2}{(t_2 + t_1)(t_3 + t_2)} = -1 \]
Thus, we can write:
\[ 4a^2 = -(t_2 + t_1)(t_3 + t_2) \quad \text{(Equation 1)} \]

### Step 3: Find the slopes of the lines AD and CD
The slope of line segment AD is given by:
\[ M_{AD} = \frac{2at_4 - 2at_1}{t_4^2 - t_1^2} = \frac{2a(t_4 - t_1)}{(t_4 - t_1)(t_4 + t_1)} = \frac{2a}{t_4 + t_1} \]

The slope of line segment CD is given by:
\[ M_{CD} = \frac{2at_4 - 2at_3}{t_4^2 - t_3^2} = \frac{2a(t_4 - t_3)}{(t_4 - t_3)(t_4 + t_3)} = \frac{2a}{t_4 + t_3} \]

### Step 4: Set up the equation for the angle subtended at point D
Since the angle subtended at point D is also \( 90^\circ \), we have:
\[ M_{AD} \cdot M_{CD} = -1 \]
Substituting the slopes:
\[ \left(\frac{2a}{t_4 + t_1}\right) \cdot \left(\frac{2a}{t_4 + t_3}\right) = -1 \]
This simplifies to:
\[ \frac{4a^2}{(t_4 + t_1)(t_4 + t_3)} = -1 \]
Thus, we can write:
\[ 4a^2 = -(t_4 + t_1)(t_4 + t_3) \quad \text{(Equation 2)} \]

### Step 5: Equate the two equations
From Equation 1 and Equation 2, we have:
\[ -(t_2 + t_1)(t_3 + t_2) = -(t_4 + t_1)(t_4 + t_3) \]
This implies:
\[ (t_2 + t_1)(t_3 + t_2) = (t_4 + t_1)(t_4 + t_3) \]

### Step 6: Solve for the required expression
We need to find the value of:
\[ \left| \frac{t_2 + t_4}{t_1 + t_3} \right| \]
From the previous equations, we can manipulate and simplify to find that:
\[ |t_2 + t_4| = |t_1 + t_3| \]
Thus:
\[ \left| \frac{t_2 + t_4}{t_1 + t_3} \right| = 1 \]

### Final Answer
\[ \boxed{1} \]