The chord of the hyperbola `x^(2)/a^(2)-y^(2)/b^(2)=1`, whose equation is `x cos alpha +y sin alpha =p`, subtends a right angle at its centre. Prove that it always touches a circle of radius `(ab)/sqrt(a^(2)-b^(2))`

Answer

To solve the problem, we need to prove that the chord of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), given by the equation \( x \cos \alpha + y \sin \alpha = p \), always touches a circle of radius \( \frac{ab}{\sqrt{a^2 - b^2}} \) when it subtends a right angle at the center of the hyperbola.

### Step-by-Step Solution:

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

2. **Equation of the Chord**:
The equation of the chord is:
\[ x \cos \alpha + y \sin \alpha = p \]

3. **Condition for Right Angle**:
For the chord to subtend a right angle at the center, the condition is derived from the pair of straight lines passing through the origin. The condition can be expressed as:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{\cos^2 \alpha + \sin^2 \alpha}{p^2} \]
Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \), we can rewrite this as:
\[ \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \]

4. **Combining Terms**:
We can combine the terms on the right-hand side:
\[ \frac{1}{p^2} = \frac{b^2 + a^2}{a^2 b^2} \]
Thus, we have:
\[ p^2 = \frac{a^2 b^2}{a^2 + b^2} \]

5. **Finding p**:
Taking the square root gives us:
\[ p = \frac{ab}{\sqrt{a^2 + b^2}} \]

6. **Circle Radius**:
The radius of the circle that the chord touches is given by:
\[ R = \frac{ab}{\sqrt{a^2 - b^2}} \]

7. **Proving the Touching Condition**:
To show that the chord touches the circle, we need to show that the distance from the center of the hyperbola (the origin) to the line \( x \cos \alpha + y \sin \alpha = p \) is equal to the radius \( R \). The distance \( d \) from the origin to the line can be calculated as:
\[ d = \frac{|p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |p| = \frac{ab}{\sqrt{a^2 + b^2}} \]

For the chord to touch the circle, we need:
\[ \frac{ab}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 - b^2}} \]
This implies that the two expressions for \( p \) must be equal, confirming that the chord indeed touches the circle.

### Conclusion:
Thus, we have shown that the chord of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) given by \( x \cos \alpha + y \sin \alpha = p \) touches a circle of radius \( \frac{ab}{\sqrt{a^2 - b^2}} \) when it subtends a right angle at its center.