If the line `y = mx + sqrt(a^(2) m^(2) -b^(2)), m = (1)/(2)` touches the hyperbola `(x^(2))/(16)-(y^(2))/(3) =1` at the point `(4 sec theta, sqrt(3) tan theta)` then `theta` is

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(2pi)/(3)`
D. `(pi)/(6)`

Answer

To solve the problem step by step, we need to find the value of \( \theta \) given the conditions of the tangent line to the hyperbola.

### Step 1: Identify the given information
We are given:
- The equation of the line:
\[ y = mx + \sqrt{a^2 m^2 - b^2} \]
where \( m = \frac{1}{2} \).
- The hyperbola:
\[ \frac{x^2}{16} - \frac{y^2}{3} = 1 \]
- The point of tangency:
\[ (4 \sec \theta, \sqrt{3} \tan \theta) \]

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at the point \( (x_1, y_1) \) is given by:
\[ \frac{x}{16} \cdot x_1 - \frac{y}{3} \cdot y_1 = 1 \]
Substituting \( x_1 = 4 \sec \theta \) and \( y_1 = \sqrt{3} \tan \theta \):
\[ \frac{x}{16} \cdot (4 \sec \theta) - \frac{y}{3} \cdot (\sqrt{3} \tan \theta) = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{4} - \frac{y \tan \theta}{\sqrt{3}} = 1 \]

### Step 3: Rearranging the tangent equation
Rearranging gives:
\[ \sec \theta \cdot x - \frac{4y \tan \theta}{\sqrt{3}} = 4 \]

### Step 4: Write the line equation with \( m = \frac{1}{2} \)
Substituting \( m = \frac{1}{2} \) into the line equation:
\[ y = \frac{1}{2}x + \sqrt{16 \cdot \left(\frac{1}{2}\right)^2 - 3} \]
Calculating the square root term:
\[ = \sqrt{16 \cdot \frac{1}{4} - 3} = \sqrt{4 - 3} = \sqrt{1} = 1 \]
Thus, the line equation becomes:
\[ y = \frac{1}{2}x + 1 \]

### Step 5: Rearranging the line equation
Rearranging gives:
\[ \frac{1}{2}x - y + 1 = 0 \]

### Step 6: Compare coefficients for tangency
Now we compare the coefficients of the two tangent equations:
1. From the hyperbola:
\[ \sec \theta = \frac{1}{2}, \quad -\frac{4 \tan \theta}{\sqrt{3}} = -1 \]

### Step 7: Solve for \( \theta \)
From \( \sec \theta = \frac{1}{2} \):
\[ \cos \theta = 2 \quad \text{(not possible)} \]
From \( -\frac{4 \tan \theta}{\sqrt{3}} = -1 \):
\[ \frac{4 \tan \theta}{\sqrt{3}} = 1 \implies \tan \theta = \frac{\sqrt{3}}{4} \]

### Step 8: Find \( \theta \)
Using \( \tan \theta = \frac{\sqrt{3}}{4} \):
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \]