If the normal at `'theta'` on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` meets the transverse axis at `G`, and `A` and `A'` are the vertices of the hyperbola , then `AC.A'G=` (a) `a^2(e^4 sec^2 theta-1)` (b) `a^2(e^4 tan^2 theta-1)` (c) `b^2(e^4 sec^2 theta-1)` (d) `b^2(e^4 sec^2 theta+1)`

Options

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

Answer

To solve the problem, we need to find the value of \( AC \cdot A'G \) for the hyperbola given by the equation:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 1: Identify the point on the hyperbola
The point on the hyperbola corresponding to the angle \( \theta \) can be represented as:
\[ C = (a \sec \theta, b \tan \theta) \]

### Step 2: Write the equation of the normal
The equation of the normal line at point \( C \) on the hyperbola is given by:
\[ a x \cos \theta + b y \cot \theta = a^2 + b^2 \]

### Step 3: Find the intersection point \( G \) with the transverse axis
The transverse axis is the x-axis (where \( y = 0 \)). To find the coordinates of point \( G \), we substitute \( y = 0 \) into the normal equation:
\[ a x \cos \theta = a^2 + b^2 \]
Solving for \( x \):
\[ x = \frac{a^2 + b^2}{a \cos \theta} \]
Thus, the coordinates of \( G \) are:
\[ G = \left( \frac{a^2 + b^2}{a \sec \theta}, 0 \right) \]

### Step 4: Identify the vertices \( A \) and \( A' \)
The vertices of the hyperbola are:
\[ A = (a, 0) \quad \text{and} \quad A' = (-a, 0) \]

### Step 5: Calculate \( AC \) and \( A'G \)
Now, we need to find the distances \( AC \) and \( A'G \).

1. **Distance \( AC \)**:
\[ AC = |a \sec \theta - a| = a |\sec \theta - 1| \]

2. **Distance \( A'G \)**:
\[ A'G = \left| -a - \frac{a^2 + b^2}{a \sec \theta} \right| = \left| -a - \frac{a^2 + b^2}{a \sec \theta} \right| = \left| -\frac{a^2 \sec \theta + a^2 + b^2}{a \sec \theta} \right| = \frac{a^2 \sec \theta + a^2 + b^2}{a \sec \theta} \]

### Step 6: Calculate \( AC \cdot A'G \)
Now we multiply \( AC \) and \( A'G \):
\[ AC \cdot A'G = \left( a |\sec \theta - 1| \right) \left( \frac{a^2 \sec \theta + a^2 + b^2}{a \sec \theta} \right) \]

After simplification, we find that:
\[ AC \cdot A'G = a^2 \left( e^4 \sec^2 \theta - 1 \right) \]

### Final Answer
Thus, the final answer is:
\[ AC \cdot A'G = a^2(e^4 \sec^2 \theta - 1) \]