The chords of contact of a point with respect to a hyperbola and its auxillary circle are perpendicular, then the point lies on :

Options

A. One of the directrices
B. one of the asymptotes
C. conjugate hyperbola
D. Axis of hyperbola

Explanation

To solve the problem, we need to find the locus of a point \( P(h, k) \) such that the chords of contact with respect to a hyperbola and its auxiliary circle are perpendicular.

### Step-by-Step Solution:

1. **Equation of the Hyperbola**:
The standard equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

2. **Equation of the Auxiliary Circle**:
The auxiliary circle of the hyperbola has the equation:
\[ x^2 + y^2 = a^2 \]

3. **Chords of Contact**:
The equation of the chord of contact from the point \( P(h, k) \) to the hyperbola is given by:
\[ \frac{hx}{a^2} - \frac{ky}{b^2} = 1 \]

The equation of the chord of contact from the point \( P(h, k) \) to the auxiliary circle is:
\[ hx + ky = a^2 \]

4. **Finding Slopes**:
- For the hyperbola, rearranging the chord of contact gives:
\[ ky = hx - a^2 \implies y = \frac{h}{k}x - \frac{a^2}{k} \]
The slope \( m_1 \) of this line is:
\[ m_1 = \frac{h}{k} \]

- For the auxiliary circle, rearranging gives:
\[ ky = a^2 - hx \implies y = -\frac{h}{k}x + \frac{a^2}{k} \]
The slope \( m_2 \) of this line is:
\[ m_2 = -\frac{h}{k} \]

5. **Condition for Perpendicularity**:
The chords are perpendicular if the product of their slopes is -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \left(\frac{h}{k}\right) \left(-\frac{h}{k}\right) = -1 \]
This simplifies to:
\[ -\frac{h^2}{k^2} = -1 \implies h^2 = k^2 \]

6. **Locus of the Point**:
From \( h^2 = k^2 \), we can write:
\[ \frac{h^2}{a^2} - \frac{k^2}{b^2} = 0 \]
Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 \]
This represents the asymptotes of the hyperbola, which can be factored as:
\[ \left(\frac{x}{a} - \frac{y}{b}\right)\left(\frac{x}{a} + \frac{y}{b}\right) = 0 \]

### Conclusion:
Thus, the locus of the point \( P(h, k) \) lies on the asymptotes of the hyperbola.