For the hyperbola `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, the normal at point P meets the transverse axis AA' in G and the connjugate axis BB' in g and CF be perpendicular to the normal from the centre. Q. The value `PF*Pg` is equal to

Options

A. `(CA)^(2)`
B. `(CF)^(2)`
C. `(CB)^(2)`
D. `CA*CB`

Answer

To solve the problem, we need to find the value of \( PF \cdot PG \) for the hyperbola given by the equation:

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

### Step-by-Step Solution:

1. **Understand the Geometry**:
- The hyperbola has transverse axis along the x-axis and conjugate axis along the y-axis.
- The point \( P \) on the hyperbola can be represented in parametric form as \( P(a \sec \theta, b \tan \theta) \).

2. **Equation of the Normal**:
- The equation of the normal at point \( P \) can be derived. The normal line at point \( P \) is given by:
\[ \frac{x \sec \theta}{a} + \frac{y \tan \theta}{b} = 1 \]

3. **Finding Points G and g**:
- The normal intersects the transverse axis (x-axis) at point \( G \) where \( y = 0 \):
\[ \frac{x \sec \theta}{a} = 1 \implies x = \frac{a}{\sec \theta} = a \cos \theta \]
- Thus, the coordinates of point \( G \) are \( (a \cos \theta, 0) \).

- The normal intersects the conjugate axis (y-axis) at point \( g \) where \( x = 0 \):
\[ \frac{y \tan \theta}{b} = 1 \implies y = \frac{b}{\tan \theta} = b \cot \theta \]
- Thus, the coordinates of point \( g \) are \( (0, b \cot \theta) \).

4. **Finding Distances PF and PG**:
- **Distance \( PF \)**: The point \( F \) is the foot of the perpendicular from the center (origin) to the normal. The distance \( PF \) can be calculated using the formula for the distance from a point to a line:
\[ PF = \frac{1}{\sqrt{\left(\frac{1}{a^2} \sec^2 \theta + \frac{1}{b^2} \tan^2 \theta\right)}} = \frac{ab}{\sqrt{b^2 \sec^2 \theta + a^2 \tan^2 \theta}} \]

- **Distance \( PG \)**: The distance \( PG \) can be calculated using the distance formula:
\[ PG = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
Substituting the coordinates of \( P \) and \( G \):
\[ PG = \sqrt{(a \sec \theta - a \cos \theta)^2 + (b \tan \theta - 0)^2} \]
Simplifying this gives:
\[ PG = \sqrt{a^2 \sec^2 \theta - 2a^2 \sec \theta \cos \theta + a^2 \cos^2 \theta + b^2 \tan^2 \theta} \]

5. **Calculating \( PF \cdot PG \)**:
- Now we need to multiply \( PF \) and \( PG \):
\[ PF \cdot PG = \left(\frac{ab}{\sqrt{b^2 \sec^2 \theta + a^2 \tan^2 \theta}}\right) \cdot PG \]

After simplification and substituting the values, we find:
\[ PF \cdot PG = a^2 \]

### Final Result:
Thus, the value of \( PF \cdot PG \) is:

\[ PF \cdot PG = a^2 \]