The locus of the trisection point of any arbitrary double ordinate of the parabola `x^(2)=4y`, is

Options

A. `9x^(2)=y`
B. `3x^(2)=2y`
C. `9x^(2)=4y`
D. `9x^(2)=2y`

Answer

To find the locus of the trisection point of any arbitrary double ordinate of the parabola \( x^2 = 4y \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Parabola**:
The given equation of the parabola is \( x^2 = 4y \). This is an upward-opening parabola with its vertex at the origin (0,0).

2. **Identifying Double Ordinates**:
A double ordinate of the parabola is a vertical line segment that intersects the parabola at two points. Let’s denote the points of intersection as \( P \) and \( Q \). For a double ordinate at \( x = 2t \) and \( x = -2t \), we can find the corresponding \( y \) values.

3. **Finding Points of Intersection**:
For the point \( Q \) at \( x = 2t \):
\[ y = \frac{(2t)^2}{4} = t^2 \]
Thus, the coordinates of point \( Q \) are \( (2t, t^2) \).

For the point \( P \) at \( x = -2t \):
\[ y = \frac{(-2t)^2}{4} = t^2 \]
Thus, the coordinates of point \( P \) are \( (-2t, t^2) \).

4. **Finding the Trisection Points**:
Let \( A \) and \( B \) be the trisection points of the segment \( PQ \). The point \( A \) divides \( PQ \) in the ratio \( 1:2 \) and point \( B \) divides \( PQ \) in the ratio \( 2:1 \).

5. **Using the Section Formula**:
The coordinates of point \( A \) can be found using the section formula:
\[ A\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right) \]
where \( m = 1 \), \( n = 2 \), \( (x_1, y_1) = (-2t, t^2) \), and \( (x_2, y_2) = (2t, t^2) \).

Substituting these values:
\[ A\left(\frac{1 \cdot 2t + 2 \cdot (-2t)}{1 + 2}, \frac{1 \cdot t^2 + 2 \cdot t^2}{1 + 2}\right) = A\left(\frac{2t - 4t}{3}, \frac{t^2 + 2t^2}{3}\right) = A\left(-\frac{2t}{3}, t^2\right) \]

6. **Expressing \( t \) in terms of \( A \)**:
Let \( A = \left(-\frac{2t}{3}, k\right) \) where \( k = t^2 \). From the x-coordinate:
\[ t = -\frac{3h}{2} \]
where \( h = -\frac{2t}{3} \).

7. **Finding \( k \)**:
Substitute \( t \) into \( k \):
\[ k = t^2 = \left(-\frac{3h}{2}\right)^2 = \frac{9h^2}{4} \]

8. **Finding the Locus**:
We have \( k = \frac{9h^2}{4} \). Replacing \( h \) with \( x \) and \( k \) with \( y \):
\[ y = \frac{9x^2}{4} \]
Rearranging gives:
\[ 9x^2 - 4y = 0 \]

### Final Answer:
The locus of the trisection point of any arbitrary double ordinate of the parabola \( x^2 = 4y \) is given by the equation:
\[ 9x^2 = 4y \]