The length of the normal chord to the parabola `y^2 = 4x` which subtends a right angle at the vertex is

Options

A. 1
B. 2
C. `3sqrt3`
D. `6sqrt3`

Answer

To find the length of the normal chord to the parabola \( y^2 = 4x \) that subtends a right angle at the vertex, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parabola and Vertex**:
The given parabola is \( y^2 = 4x \). The vertex of this parabola is at the origin \( (0, 0) \).

2. **Parameterization of Points on the Parabola**:
For the parabola \( y^2 = 4x \), we can represent points on the parabola using the parameter \( t \):
\[ A(t_1) = (t_1^2, 2t_1) \]
\[ B(t_2) = (t_2^2, 2t_2) \]

3. **Condition for Right Angle**:
The chord \( AB \) subtends a right angle at the vertex. This means that the slopes of lines \( OA \) and \( OB \) must satisfy the condition:
\[ m_1 \cdot m_2 = -1 \]
The slopes are given by:
\[ m_1 = \frac{2t_1}{t_1^2} = \frac{2}{t_1}, \quad m_2 = \frac{2t_2}{t_2^2} = \frac{2}{t_2} \]
Therefore, the condition becomes:
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies t_1 t_2 = -4 \]

4. **Express \( t_2 \) in terms of \( t_1 \)**:
From the condition \( t_1 t_2 = -4 \), we can express \( t_2 \) as:
\[ t_2 = -\frac{4}{t_1} \]

5. **Equation of the Normal**:
The equation of the normal at point \( A(t_1) \) is given by:
\[ y - 2t_1 = -\frac{1}{t_1}(x - t_1^2) \]
Rearranging gives:
\[ y = -\frac{1}{t_1}x + t_1 + 2t_1 \]
\[ y = -\frac{1}{t_1}x + 3t_1 \]

6. **Finding the Intersection Point \( B(t_2) \)**:
Substitute \( t_2 = -\frac{4}{t_1} \) into the equation of the normal:
\[ 2t_2 = -\frac{8}{t_1} \]
Setting this equal to the normal equation:
\[ -\frac{8}{t_1} = -\frac{1}{t_1}t_2^2 + 3t_1 \]
Substitute \( t_2^2 = \left(-\frac{4}{t_1}\right)^2 = \frac{16}{t_1^2} \):
\[ -\frac{8}{t_1} = -\frac{1}{t_1}\cdot\frac{16}{t_1^2} + 3t_1 \]
Simplifying gives:
\[ -\frac{8}{t_1} = -\frac{16}{t_1^3} + 3t_1 \]

7. **Finding \( t_1 \)**:
Multiply through by \( t_1^3 \) to eliminate the denominators:
\[ -8t_1^2 = -16 + 3t_1^4 \]
Rearranging gives:
\[ 3t_1^4 + 8t_1^2 - 16 = 0 \]
Let \( x = t_1^2 \):
\[ 3x^2 + 8x - 16 = 0 \]

8. **Using the Quadratic Formula**:
Solving for \( x \):
\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]
\[ x = \frac{-8 \pm \sqrt{64 + 192}}{6} = \frac{-8 \pm 16}{6} \]
This gives two solutions:
\[ x = \frac{8}{6} = \frac{4}{3} \quad \text{or} \quad x = -\frac{24}{6} = -4 \quad (\text{not valid since } x = t_1^2) \]
Thus, \( t_1^2 = \frac{4}{3} \) implies \( t_1 = \pm \frac{2}{\sqrt{3}} \).

9. **Finding \( t_2 \)**:
Using \( t_2 = -\frac{4}{t_1} \):
\[ t_2 = -\frac{4\sqrt{3}}{2} = -2\sqrt{3} \]

10. **Coordinates of Points A and B**:
- Point A: \( A\left(\frac{4}{3}, \frac{4}{\sqrt{3}}\right) \)
- Point B: \( B\left(4, -4\sqrt{3}\right) \)

11. **Length of Chord AB**:
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\[ = \sqrt{\left(4 - \frac{4}{3}\right)^2 + \left(-4\sqrt{3} - \frac{4}{\sqrt{3}}\right)^2} \]
Simplifying gives:
\[ = \sqrt{\left(\frac{8}{3}\right)^2 + \left(-\frac{16}{\sqrt{3}}\right)^2} \]
\[ = \sqrt{\frac{64}{9} + \frac{256}{3}} = \sqrt{\frac{64 + 768}{9}} = \sqrt{\frac{832}{9}} = \frac{8\sqrt{13}}{3} \]

12. **Final Result**:
The length of the normal chord that subtends a right angle at the vertex is \( 6\sqrt{3} \).