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 that the normals at two points on the parabola \(y^2 = 4x\) meet again on the parabola and that \(x_1 + x_2 = 8\).

### Step-by-step Solution:

1. **Identify the points on the parabola:**
The points \((x_1, y_1)\) and \((x_2, y_2)\) can be represented in terms of parameters \(t_1\) and \(t_2\):
\[ (x_1, y_1) = (t_1^2, 2t_1) \quad \text{and} \quad (x_2, y_2) = (t_2^2, 2t_2) \]

2. **Use the given condition \(x_1 + x_2 = 8\):**
From the problem, we know:
\[ t_1^2 + t_2^2 = 8 \]

3. **Find the equations of the normals:**
The equation of the normal at point \((t_1^2, 2t_1)\) is:
\[ y - 2t_1 = -\frac{1}{t_1}(x - t_1^2) \]
Rearranging gives:
\[ y = -\frac{1}{t_1}x + \left(2t_1 + \frac{t_1^2}{t_1}\right) = -\frac{1}{t_1}x + 3t_1 \]

The equation of the normal at point \((t_2^2, 2t_2)\) is:
\[ y - 2t_2 = -\frac{1}{t_2}(x - t_2^2) \]
Rearranging gives:
\[ y = -\frac{1}{t_2}x + 3t_2 \]

4. **Find the intersection of the normals:**
Set the equations of the normals equal to each other:
\[ -\frac{1}{t_1}x + 3t_1 = -\frac{1}{t_2}x + 3t_2 \]
Rearranging gives:
\[ \left(\frac{1}{t_1} - \frac{1}{t_2}\right)x = 3(t_2 - t_1) \]
Thus,
\[ x = \frac{3(t_2 - t_1)}{\frac{1}{t_1} - \frac{1}{t_2}} = \frac{3(t_2 - t_1)t_1t_2}{t_2 - t_1} = 3t_1t_2 \]

5. **Find the corresponding y-coordinate:**
Substitute \(x = 3t_1t_2\) into one of the normal equations:
\[ y = -\frac{1}{t_1}(3t_1t_2) + 3t_1 = 3t_1 - 3t_2 = 3(t_1 - t_2) \]

6. **Calculate \(|y_1 - y_2|\):**
We know:
\[ |y_1 - y_2| = |2t_1 - 2t_2| = 2|t_1 - t_2| \]

7. **Find \(t_1^2 + t_2^2\) and \(t_1t_2\):**
From the equations, we have:
\[ (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2 \]
Let \(s = t_1 + t_2\) and \(p = t_1t_2\). We know:
\[ t_1^2 + t_2^2 = 8 \quad \text{and} \quad t_1t_2 = 2 \]
Therefore:
\[ s^2 = 8 + 2 \cdot 2 = 12 \quad \Rightarrow \quad s = 2\sqrt{3} \]

8. **Calculate \(|t_1 - t_2|\):**
We can find \(|t_1 - t_2|\) using:
\[ |t_1 - t_2| = \sqrt{(t_1 + t_2)^2 - 4t_1t_2} = \sqrt{12 - 8} = \sqrt{4} = 2 \]

9. **Final calculation of \(|y_1 - y_2|\):**
Thus,
\[ |y_1 - y_2| = 2|t_1 - t_2| = 2 \cdot 2 = 4 \]

### Conclusion:
The final answer is:
\[ |y_1 - y_2| = 4 \]