If normals at the ends of the double ordinate x = 4 of parabola `y^(2)=4x` meet the curve again in P and P' respectively, then PP' =

Options

A. 6
B. 12
C. 10
D. none of these

Explanation

To solve the problem, we need to find the length of the line segment \( PP' \) where \( P \) and \( P' \) are the points where the normals at the ends of the double ordinate \( x = 4 \) of the parabola \( y^2 = 4x \) meet the curve again.

### Step-by-Step Solution:

1. **Identify the given parabola**:
The equation of the parabola is \( y^2 = 4x \). This can be compared to the standard form \( y^2 = 4ax \) where \( a = 1 \).

2. **Find the points on the double ordinate \( x = 4 \)**:
Substitute \( x = 4 \) into the parabola's equation:
\[ y^2 = 4(4) = 16 \implies y = \pm 4. \]
Therefore, the points on the double ordinate are \( A(4, 4) \) and \( B(4, -4) \).

3. **Find the parameters for points A and B**:
For point \( A \) (where \( y = 4 \)):
\[ y = 2t_1 \implies 2t_1 = 4 \implies t_1 = 2. \]
For point \( B \) (where \( y = -4 \)):
\[ y = 2t_2 \implies 2t_2 = -4 \implies t_2 = -2. \]

4. **Find the equations of the normals at points A and B**:
The equation of the normal to the parabola at a point \( (t^2, 2t) \) is given by:
\[ y - 2t = -\frac{1}{t}(x - t^2). \]
For point \( A(4, 4) \) with \( t_1 = 2 \):
\[ y - 4 = -\frac{1}{2}(x - 4) \implies 2y - 8 = -x + 4 \implies x + 2y = 12. \]
For point \( B(4, -4) \) with \( t_2 = -2 \):
\[ y + 4 = \frac{1}{2}(x - 4) \implies 2y + 8 = x - 4 \implies x - 2y = 12. \]

5. **Find the intersection points \( P \) and \( P' \)**:
To find \( P \), substitute \( x = 12 - 2y \) into the parabola's equation:
\[ y^2 = 4(12 - 2y) \implies y^2 + 8y - 48 = 0. \]
Solving this quadratic equation using the quadratic formula:
\[ y = \frac{-8 \pm \sqrt{8^2 + 4 \cdot 48}}{2} = \frac{-8 \pm \sqrt{64 + 192}}{2} = \frac{-8 \pm 16}{2}. \]
This gives \( y = 4 \) and \( y = -12 \). The corresponding \( x \) values are:
For \( y = 4 \):
\[ x = 12 - 2(4) = 4 \quad \text{(point A)}. \]
For \( y = -12 \):
\[ x = 12 - 2(-12) = 36 \quad \text{(point P)}. \]
Thus, \( P(36, -12) \).

Now, for \( P' \), substitute \( x = 12 + 2y \) into the parabola's equation:
\[ y^2 = 4(12 + 2y) \implies y^2 - 8y - 48 = 0. \]
Solving this quadratic gives:
\[ y = \frac{8 \pm \sqrt{64 + 192}}{2} = \frac{8 \pm 16}{2}. \]
This gives \( y = 12 \) and \( y = -4 \). The corresponding \( x \) values are:
For \( y = 12 \):
\[ x = 12 + 2(12) = 36 \quad \text{(point P')}. \]
For \( y = -4 \):
\[ x = 12 + 2(-4) = 4 \quad \text{(point B)}. \]
Thus, \( P'(36, 12) \).

6. **Calculate the distance \( PP' \)**:
The distance between points \( P(36, -12) \) and \( P'(36, 12) \) is:
\[ PP' = \sqrt{(36 - 36)^2 + (12 - (-12))^2} = \sqrt{0 + 24^2} = 24. \]

### Final Answer:
Thus, the length \( PP' = 24 \).