A series of chords of the hyperbola `x^(2)//a^(2)-y^(2)//b^(2)=1` touch the circle on the line joining the foci as diameter. The locus of the poles of these chords with respect to the hyperbola is
`(x^(2))/(a^4)+(y^(2))/(b^(4))=(1)/(a^(2)+b^(2))`

Answer

To solve the problem, we need to find the locus of the poles of the chords of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) that touch the circle whose diameter is the line joining the foci of the hyperbola.

### Step-by-Step Solution:

1. **Understanding the Hyperbola and Circle**:
The hyperbola given is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). The foci of this hyperbola are located at \( (c, 0) \) and \( (-c, 0) \), where \( c = \sqrt{a^2 + b^2} \). The circle with diameter along the line joining the foci has its center at the origin and radius \( \frac{c}{2} \).

2. **Equation of the Circle**:
The equation of the circle is given by:
\[ x^2 + y^2 = \left(\frac{c}{2}\right)^2 = \frac{c^2}{4} = \frac{a^2 + b^2}{4} \]

3. **Chords of the Hyperbola**:
Let \( P(a \sec \theta, b \tan \theta) \) and \( Q(a \sec \phi, b \tan \phi) \) be two points on the hyperbola. The equation of the chord joining these two points can be derived using the two-point form of the line.

4. **Normal to the Hyperbola**:
The normal at point \( P \) can be expressed as:
\[ \frac{x}{a \cos \theta} - \frac{y}{b \sin \theta} = \frac{1}{\cos \theta} \]

5. **Distance from the Circle**:
The distance \( D \) from the center of the circle to the line representing the chord must equal the radius of the circle for the chord to touch the circle. The distance \( D \) from the origin to the line can be calculated using the formula for the distance from a point to a line.

6. **Setting Up the Equation**:
After finding the distance, we set it equal to \( \frac{c}{2} \) and simplify the resulting equation. This will yield a relationship involving \( \theta \) and \( \phi \).

7. **Finding the Locus**:
By eliminating \( \theta \) and \( \phi \) from the equations, we can find the locus of the poles of these chords. The locus will be in the form of an equation involving \( x \) and \( y \).

8. **Final Equation**:
After simplification, we arrive at the equation:
\[ \frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2 + b^2} \]