OA is the chord of the parabola `y^(2)=4x` and perpendicular to OA which cuts the axis of the parabola at C. If the foot of A on the axis of the parabola is D, then the length CD is equal to

Options

A. 2 units
B. 3 units
C. 4 units
D. 6 units

Answer

To solve the problem, we need to find the length \( CD \) where \( C \) is the point where the line perpendicular to the chord \( OA \) intersects the axis of the parabola, and \( D \) is the foot of point \( A \) on the axis of the parabola.

### Step-by-step Solution:

1. **Identify the parabola and its properties**:
The equation of the parabola is given by \( y^2 = 4x \). This is a standard parabola that opens to the right.

2. **General point on the parabola**:
A general point \( A \) on the parabola can be represented as \( (t^2, 2at) \). For our parabola, \( a = 1 \), so the point \( A \) can be represented as \( (t^2, 2t) \).

3. **Find the slope of the chord \( OA \)**:
The slope of the line \( OA \) can be calculated using the coordinates of point \( A \) and the origin \( O(0, 0) \):
\[ \text{slope of } OA = \frac{2t - 0}{t^2 - 0} = \frac{2t}{t^2} = \frac{2}{t} \]

4. **Determine the slope of the perpendicular line \( AB \)**:
The slope of the line perpendicular to \( OA \) is the negative reciprocal of the slope of \( OA \):
\[ \text{slope of } AB = -\frac{t}{2} \]

5. **Write the equation of line \( AB \)**:
Using the point-slope form of the equation of a line, the equation of line \( AB \) passing through point \( A(t^2, 2t) \) with slope \( -\frac{t}{2} \) is:
\[ y - 2t = -\frac{t}{2}(x - t^2) \]

6. **Find the intersection point \( C \) with the x-axis**:
To find point \( C \), set \( y = 0 \) in the equation of line \( AB \):
\[ 0 - 2t = -\frac{t}{2}(x - t^2) \]
Simplifying this gives:
\[ -2t = -\frac{t}{2}(x - t^2) \]
Multiplying through by -2 gives:
\[ 4t = t(x - t^2) \]
Dividing by \( t \) (assuming \( t \neq 0 \)):
\[ 4 = x - t^2 \implies x = t^2 + 4 \]
Therefore, point \( C \) is \( (t^2 + 4, 0) \).

7. **Identify point \( D \)**:
The foot of point \( A \) on the axis of the parabola is simply the projection of \( A \) onto the x-axis, which is \( D(t^2, 0) \).

8. **Calculate the length \( CD \)**:
The length \( CD \) can be calculated as the difference in the x-coordinates of points \( C \) and \( D \):
\[ CD = |(t^2 + 4) - t^2| = |4| = 4 \]

Thus, the length \( CD \) is equal to **4**.