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)/((CB^(2)))` is equal to

Options

A. (a)`4`
B. (b)`3`
C. (c)`2`
D. (d)`1`

Answer

To solve the problem, we need to find the value of \((PF \cdot PG) / (CB^2)\) for the hyperbola given by the equation:

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

### Step-by-Step Solution

1. **Understanding the Hyperbola**:
The hyperbola is centered at the origin (0, 0) with the transverse axis along the x-axis and the conjugate axis along the y-axis. The vertices are at \((\pm a, 0)\) and the asymptotes are given by the equations \(y = \pm \frac{b}{a} x\).

2. **Coordinates of Point P**:
Let the coordinates of point \(P\) on the hyperbola be given by:
\[ P(a \sec \theta, b \tan \theta) \]
where \(\theta\) is the angle corresponding to point \(P\).

3. **Equation of the Normal at Point P**:
The equation of the normal line at point \(P\) can be derived from the slope of the tangent line at \(P\). The slope of the tangent at point \(P\) is given by:
\[ \frac{dy}{dx} = \frac{b^2 x}{a^2 y} \]
Therefore, the slope of the normal is:
\[ -\frac{a^2 y}{b^2 x} \]

The equation of the normal line can be written as:
\[ y - b \tan \theta = -\frac{a^2 b \tan \theta}{b^2 a \sec \theta}(x - a \sec \theta) \]

4. **Finding Points G and g**:
- **Point G**: The normal meets the transverse axis (x-axis) at point \(G\). Setting \(y = 0\) in the normal equation gives the x-coordinate of \(G\).
- **Point g**: The normal meets the conjugate axis (y-axis) at point \(g\). Setting \(x = 0\) in the normal equation gives the y-coordinate of \(g\).

5. **Distance PF**:
The distance \(PF\) is the perpendicular distance from point \(P\) to the normal line. This can be calculated using the formula for the distance from a point to a line.

6. **Distance PG**:
The distance \(PG\) can be calculated as the distance from point \(P\) to point \(G\).

7. **Calculating CB**:
The distance \(CB\) is the distance from the center \(C(0, 0)\) to point \(B(0, b)\), which is simply \(b\).

8. **Final Calculation**:
We need to calculate:
\[ \frac{PF \cdot PG}{CB^2} \]
After substituting the distances calculated in the previous steps, we simplify the expression.

9. **Conclusion**:
After performing the calculations, we find that:
\[ \frac{PF \cdot PG}{CB^2} = 1 \]

### Final Answer:
The value of \(\frac{PF \cdot PG}{CB^2}\) is equal to \(1\).