If the line `y = mx + sqrt(a^(2)m^(2) - b^(2)) ` touches the hyperbola `x^(2)/a^(2) - y^(2)/b^(2) = 1` at the point `(a sec theta, b tan theta)`, then find `theta` .

Answer

To solve the problem, we need to find the angle \( \theta \) given that the line \( y = mx + \sqrt{a^2 m^2 - b^2} \) touches the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( (a \sec \theta, b \tan \theta) \).

### Step-by-Step Solution:

1. **Identify the Point on the Hyperbola:**
The point \( (a \sec \theta, b \tan \theta) \) must satisfy the hyperbola equation:
\[ \frac{(a \sec \theta)^2}{a^2} - \frac{(b \tan \theta)^2}{b^2} = 1 \]
Simplifying this gives:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
This is a known trigonometric identity, confirming that the point lies on the hyperbola.

2. **Equation of the Tangent to the Hyperbola:**
The equation of the tangent to the hyperbola at the point \( (a \sec \theta, b \tan \theta) \) can be expressed as:
\[ \frac{a \sec \theta}{a^2} x - \frac{b \tan \theta}{b^2} y = 1 \]
Simplifying this gives:
\[ \sec \theta \cdot x - \frac{\tan \theta}{b} \cdot y = 1 \]

3. **Rearranging the Tangent Equation:**
Rearranging the equation gives:
\[ y = \frac{b \tan \theta}{a \sec \theta} x - b \]
This can be rewritten as:
\[ y = \frac{b}{a \sin \theta} x - b \cot \theta \]

4. **Comparing with the Given Line:**
The line given is:
\[ y = mx + \sqrt{a^2 m^2 - b^2} \]
For the two equations to represent the same line, their slopes must be equal:
\[ m = \frac{b}{a \sin \theta} \]
And their y-intercepts must also be equal:
\[ \sqrt{a^2 m^2 - b^2} = -b \cot \theta \]

5. **Finding \( \theta \):**
From the slope equation:
\[ \sin \theta = \frac{b}{a m} \]
Taking the inverse sine gives:
\[ \theta = \sin^{-1}\left(\frac{b}{a m}\right) \]

### Final Answer:
Thus, the value of \( \theta \) is:
\[ \theta = \sin^{-1}\left(\frac{b}{a m}\right) \]