The locus of the poles of normal chords of the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`, is

Options

A. `x^2y^2(a^2-b^2)^2=a^6y^2+b^6x^2`
B. `x^2y^2(a^2+b^2)^2=a^6y^2+b^6x^2`
C. `2x^2y^2(a^2-b^2)^2=a^6y^2+b^6x^2`
D. `2x^2y^2(a^2+b^2)^2=a^6y^2+b^6x^2`

Answer

To find the locus of the poles of normal chords of the ellipse given by the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

we can follow these steps:

### Step 1: Define the Polar of a Point
Let \( (h, k) \) be the coordinates of the pole. The polar of this point with respect to the ellipse is given by the equation:

\[ \frac{hx}{a^2} + \frac{ky}{b^2} = 1. \]

### Step 2: Equation of the Normal
The equation of the normal to the ellipse at the point \( (x_0, y_0) \) can be expressed as:

\[ \frac{a^2}{b^2} (y - y_0) = -\frac{b^2}{a^2} (x - x_0). \]

### Step 3: Relate the Normal to the Polar
For the normal to be a chord of the ellipse, we need to relate the coefficients of the normal equation to the polar equation. The normal can be expressed in terms of the angle \( \theta \) as:

\[ b^2 y \cos \theta - a^2 x \sin \theta = a^2 - b^2. \]

### Step 4: Comparing Coefficients
We can compare the coefficients of the normal and the polar equations. From the polar equation, we have:

\[ \frac{h}{a^2} = \frac{k}{b^2} = \frac{1}{a^2 - b^2}. \]

### Step 5: Express Cosine and Sine
From the above relationships, we can express \( \cos \theta \) and \( \sin \theta \):

\[ \cos \theta = \frac{e^3}{h(a^2 - b^2)}, \]
\[ \sin \theta = -\frac{b^3}{k(a^2 - b^2)}. \]

### Step 6: Use the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \left(\frac{e^3}{h(a^2 - b^2)}\right)^2 + \left(-\frac{b^3}{k(a^2 - b^2)}\right)^2 = 1. \]

### Step 7: Simplify the Equation
Squaring both sides and simplifying, we get:

\[ \frac{e^6}{h^2(a^2 - b^2)^2} + \frac{b^6}{k^2(a^2 - b^2)^2} = 1. \]

### Step 8: Rearranging the Terms
Multiplying through by \( h^2 k^2 (a^2 - b^2)^2 \):

\[ e^6 k^2 + b^6 h^2 = h^2 k^2 (a^2 - b^2)^2. \]

### Step 9: Final Equation
This leads to the locus of the poles of the normal chords of the ellipse, which can be expressed in the form:

\[ \frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2. \]

### Conclusion
Thus, the locus of the poles of normal chords of the ellipse is given by:

\[ \frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2. \]