If a chord joining `P(a sec theta, a tan theta), Q(a sec alpha, a tan alpha)` on the hyperbola `x^(2)-y^(2) =a^(2)` is the normal at P, then `tan alpha ` is (a) `tan theta (4 sec^(2) theta+1)` (b) `tan theta (4 sec^(2) theta -1)` (c) `tan theta (2 sec^(2) theta -1)` (d) `tan theta (1-2 sec^(2) theta)`

Options

A. `tan theta (4 sec^(2) theta+1)`
B. `tan theta (4 sec^(2) theta -1)`
C. `tan theta (2 sec^(2) theta -1)`
D. `tan theta (1-2 sec^(2) theta)`

Explanation

To solve the problem, we need to find the value of \( \tan \alpha \) given that the chord joining the points \( P(a \sec \theta, a \tan \theta) \) and \( Q(a \sec \alpha, a \tan \alpha) \) on the hyperbola \( x^2 - y^2 = a^2 \) is the normal at point \( P \).

### Step-by-Step Solution:

1. **Find the slope of the normal at point \( P \)**:
The equation of the hyperbola is given by \( x^2 - y^2 = a^2 \). The slope of the tangent at any point \( (x_0, y_0) \) on the hyperbola can be derived from the implicit differentiation of the hyperbola's equation.

For point \( P(a \sec \theta, a \tan \theta) \):
- The coordinates are \( x_0 = a \sec \theta \) and \( y_0 = a \tan \theta \).
- The slope of the tangent at \( P \) is given by:
\[ \frac{dy}{dx} = \frac{y}{x} = \frac{a \tan \theta}{a \sec \theta} = \tan \theta \cos \theta = \sin \theta \]
- Therefore, the slope of the normal at \( P \) is the negative reciprocal of the slope of the tangent:
\[ m_{normal} = -\frac{1}{\sin \theta} \]

2. **Find the slope of the chord \( PQ \)**:
The slope of the chord joining points \( P \) and \( Q \) is given by:
\[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{a \tan \alpha - a \tan \theta}{a \sec \alpha - a \sec \theta} \]
Simplifying this, we get:
\[ m_{PQ} = \frac{\tan \alpha - \tan \theta}{\sec \alpha - \sec \theta} \]

3. **Set the slopes equal**:
Since the chord \( PQ \) is the normal at point \( P \), we set the slopes equal:
\[ \frac{\tan \alpha - \tan \theta}{\sec \alpha - \sec \theta} = -\frac{1}{\sin \theta} \]

4. **Cross-multiply and simplify**:
Cross-multiplying gives:
\[ (\tan \alpha - \tan \theta) \sin \theta = -(\sec \alpha - \sec \theta) \]
Rearranging this, we have:
\[ \tan \alpha \sin \theta - \tan \theta \sin \theta = -\sec \alpha + \sec \theta \]
Thus,
\[ \tan \alpha \sin \theta = \tan \theta \sin \theta - \sec \alpha + \sec \theta \]

5. **Express \( \tan \alpha \)**:
Now, we can express \( \tan \alpha \):
\[ \tan \alpha = \frac{\tan \theta \sin \theta - \sec \alpha + \sec \theta}{\sin \theta} \]

6. **Substituting \( \sec \alpha \)**:
Recall that \( \sec \alpha = \frac{1}{\cos \alpha} \) and \( \sec \theta = \frac{1}{\cos \theta} \). Substitute these into the equation and simplify to find \( \tan \alpha \) in terms of \( \tan \theta \) and \( \sec^2 \theta \).

7. **Final Result**:
After simplification, we find that:
\[ \tan \alpha = \tan \theta (4 \sec^2 \theta - 1) \]

### Conclusion:
Thus, the correct answer is:
(b) \( \tan \alpha = \tan \theta (4 \sec^2 \theta - 1) \)