If the normals at two points `(x_(1),y_(1))` and `(x_(2),y_(2))` of the parabola `y^(2)=4x` meets again on the parabola, where `x_(1)+x_(2)=8` then `|y_(1)-y_(2)|` is equal to

Options

A. `sqrt2`
B. 3
C. 4
D. 2

Answer

To solve the problem, we need to find the value of \(|y_1 - y_2|\) given the conditions of the parabola \(y^2 = 4x\) and the points \((x_1, y_1)\) and \((x_2, y_2)\) where \(x_1 + x_2 = 8\).

### Step-by-Step Solution:

1. **Identify the points on the parabola**:
The points on the parabola \(y^2 = 4x\) can be expressed in terms of a parameter \(t\):
\[ (x_1, y_1) = (t_1^2, 2t_1) \quad \text{and} \quad (x_2, y_2) = (t_2^2, 2t_2) \]

2. **Use the condition \(x_1 + x_2 = 8\)**:
From the points we defined:
\[ t_1^2 + t_2^2 = 8 \]

3. **Find the equations of the normals**:
The equation of the normal to the parabola at point \((t^2, 2t)\) is given by:
\[ y - 2t = -\frac{1}{t}(x - t^2) \]
Rearranging gives:
\[ y = -\frac{1}{t}x + \left(2t + \frac{t^2}{t}\right) = -\frac{1}{t}x + 3t \]
Thus, the normals at points \((x_1, y_1)\) and \((x_2, y_2)\) are:
\[ y = -\frac{1}{t_1}x + 3t_1 \quad \text{and} \quad y = -\frac{1}{t_2}x + 3t_2 \]

4. **Find the intersection of the normals**:
Set the two equations equal to find the intersection point:
\[ -\frac{1}{t_1}x + 3t_1 = -\frac{1}{t_2}x + 3t_2 \]
Rearranging gives:
\[ \left(\frac{1}{t_1} - \frac{1}{t_2}\right)x = 3(t_1 - t_2) \]
Thus, we can solve for \(x\):
\[ x = \frac{3(t_1 - t_2)}{\frac{1}{t_1} - \frac{1}{t_2}} = 3(t_1 - t_2) \cdot \frac{t_1 t_2}{t_2 - t_1} = -3t_1 t_2 \]

5. **Substituting back to find \(y\)**:
Substitute \(x = -3t_1 t_2\) into either normal equation to find \(y\):
\[ y = -\frac{1}{t_1}(-3t_1 t_2) + 3t_1 = 3t_2 + 3t_1 = 3(t_1 + t_2) \]

6. **Relate \(y_1\) and \(y_2\)**:
Now, we can express \(|y_1 - y_2|\):
\[ |y_1 - y_2| = |2t_1 - 2t_2| = 2|t_1 - t_2| \]

7. **Find \(t_1 - t_2\)**:
From the earlier equations, we have:
\[ t_1^2 + t_2^2 = 8 \quad \text{and} \quad t_1 + t_2 = 2\sqrt{3} \]
Using the identity \((t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2\):
\[ (2\sqrt{3})^2 = 8 + 2t_1t_2 \implies 12 = 8 + 2t_1t_2 \implies 2t_1t_2 = 4 \implies t_1t_2 = 2 \]

8. **Solve for \(t_1 - t_2\)**:
Using the identity:
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2 \implies 8 = 12 - 4 \implies t_1^2 + t_2^2 = 8 \]
Thus, we can solve for \(t_1 - t_2\):
\[ (t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 = 12 - 8 = 4 \implies t_1 - t_2 = \pm 2 \]

9. **Final Calculation**:
Thus, we have:
\[ |y_1 - y_2| = 2|t_1 - t_2| = 2 \cdot 2 = 4 \]

### Conclusion:
The value of \(|y_1 - y_2|\) is \(4\).