A point P is taken on the circle `x^(2)+y^(2)=a^(2)` and PN, PM are draw, perpendicular to the axes. The locus of the pole of the line MN is

Options

A. `x^(2)+y^(2)=a^(2)`
B. `x^(-2)+y^(-2)=2x^(-2)`
C. `x^(-2)+y^(-2)=a^(-2)`
D. `x^(-2)+y^(-2)=a^(-4)`

Answer

To solve the problem, we need to find the locus of the pole of the line MN, where M and N are points on the axes corresponding to a point P on the circle defined by the equation \(x^2 + y^2 = a^2\).

### Step-by-Step Solution:

1. **Identify the Circle and Point P**:
- The circle is given by the equation \(x^2 + y^2 = a^2\).
- Let point \(P\) on the circle be represented in parametric form as \(P(a \cos \theta, a \sin \theta)\), where \(\theta\) is the angle parameter.

2. **Determine Points M and N**:
- Point \(M\) is the projection of \(P\) on the y-axis, which gives us the coordinates \(M(0, a \sin \theta)\).
- Point \(N\) is the projection of \(P\) on the x-axis, which gives us the coordinates \(N(a \cos \theta, 0)\).

3. **Equation of Line MN**:
- The slope of line \(MN\) can be calculated using the coordinates of points \(M\) and \(N\):
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - a \sin \theta}{a \cos \theta - 0} = -\frac{a \sin \theta}{a \cos \theta} = -\tan \theta \]
- Using point-slope form, the equation of line \(MN\) is:
\[ y - a \sin \theta = -\tan \theta (x - 0) \]
Simplifying this gives:
\[ y = -\tan \theta \cdot x + a \sin \theta \]

4. **Finding the Pole of Line MN**:
- The pole of the line \(MN\) with respect to the circle \(x^2 + y^2 = a^2\) can be found using the formula for the chord of contact:
\[ \frac{x}{h} + \frac{y}{k} = 1 \]
- Here, \(h\) and \(k\) are the x-intercept and y-intercept of the line, respectively. From the equation derived, we can express:
\[ h = \frac{a^2}{a \cos \theta} = \frac{a}{\cos \theta}, \quad k = \frac{a^2}{a \sin \theta} = \frac{a}{\sin \theta} \]

5. **Using Trigonometric Identity**:
- From the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can substitute for \(h\) and \(k\):
\[ \left(\frac{a}{k}\right)^2 + \left(\frac{a}{h}\right)^2 = 1 \]
- This leads to:
\[ \frac{a^2}{k^2} + \frac{a^2}{h^2} = 1 \]

6. **Final Locus Equation**:
- Replacing \(h\) with \(x\) and \(k\) with \(y\), we get:
\[ \frac{a^2}{y^2} + \frac{a^2}{x^2} = 1 \]
- Rearranging gives:
\[ \frac{1}{y^2} + \frac{1}{x^2} = \frac{1}{a^2} \]

### Final Answer:
The locus of the pole of the line \(MN\) is given by:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{a^2} \]