Q is any point on the circle `x^(2) +y^(2) = 9. QN` is perpendicular from Q to the x-axis. Locus of the point of trisection of QN is

Options

A. `4x^(2) +9y^(2) = 36`
B. `9x^(2) +4y^(2) = 36`
C. `9x^(2) +y^(2) = 9`
D. `x^(2) +9y^(2) = 9`

Explanation

To solve the problem, we need to find the locus of the points of trisection of the perpendicular segment \( QN \) from a point \( Q \) on the circle defined by the equation \( x^2 + y^2 = 9 \) to the x-axis.

### Step-by-Step Solution:

1. **Identify the Circle and Point \( Q \)**:
The equation of the circle is given by:
\[ x^2 + y^2 = 9 \]
This represents a circle with center at the origin (0, 0) and radius \( r = 3 \).

2. **General Coordinates of Point \( Q \)**:
A point \( Q \) on the circle can be represented in parametric form as:
\[ Q = (3 \cos \theta, 3 \sin \theta) \]
where \( \theta \) is the angle parameter.

3. **Determine the Coordinates of Point \( N \)**:
Since \( QN \) is perpendicular to the x-axis, the coordinates of point \( N \) (the foot of the perpendicular from \( Q \) to the x-axis) are:
\[ N = (3 \cos \theta, 0) \]

4. **Finding Points of Trisection**:
The segment \( QN \) can be divided into three equal parts. The points of trisection can be found as follows:
- The first point of trisection \( P_1 \) is located at:
\[ P_1 = \left(3 \cos \theta, \frac{1}{3}(3 \sin \theta)\right) = \left(3 \cos \theta, \sin \theta\right) \]
- The second point of trisection \( P_2 \) is located at:
\[ P_2 = \left(3 \cos \theta, \frac{2}{3}(3 \sin \theta)\right) = \left(3 \cos \theta, 2 \sin \theta\right) \]

5. **Finding the Locus of Points**:
We will find the locus of both points \( P_1 \) and \( P_2 \).

- For point \( P_1 \):
\[ x = 3 \cos \theta \quad \text{and} \quad y = \sin \theta \]
From \( x = 3 \cos \theta \), we have:
\[ \cos \theta = \frac{x}{3} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ y^2 + \left(\frac{x}{3}\right)^2 = 1 \implies y^2 + \frac{x^2}{9} = 1 \]
Multiplying through by 9 gives:
\[ 9y^2 + x^2 = 9 \]

- For point \( P_2 \):
\[ x = 3 \cos \theta \quad \text{and} \quad y = 2 \sin \theta \]
From \( x = 3 \cos \theta \), we again have:
\[ \cos \theta = \frac{x}{3} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{y}{2}\right)^2 + \left(\frac{x}{3}\right)^2 = 1 \implies \frac{y^2}{4} + \frac{x^2}{9} = 1 \]
Multiplying through by 36 gives:
\[ 4y^2 + 9x^2 = 36 \]

6. **Final Locus Equations**:
The locus of the points of trisection are given by:
- For \( P_1 \): \( 9y^2 + x^2 = 9 \)
- For \( P_2 \): \( 4y^2 + 9x^2 = 36 \)