Normal is drawn at one of the extremities of the latus rectum of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` which meets the axes at points `Aa n dB` . Then find the area of triangle `O A B(O` being the origin).

Explanation

To solve the problem, we need to find the area of triangle OAB where O is the origin, A is the point where the normal to the hyperbola meets the x-axis, and B is the point where the normal meets the y-axis.

### Step-by-Step Solution:

1. **Identify the Hyperbola**:
The given hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

2. **Find the Extremities of the Latus Rectum**:
The extremities of the latus rectum of the hyperbola are given by the points \((a, \frac{b^2}{a})\) and \((-a, \frac{b^2}{a})\). We will consider the point \(P(a, \frac{b^2}{a})\).

3. **Equation of the Normal**:
The equation of the normal to the hyperbola at point \(P(x_1, y_1) = (a, \frac{b^2}{a})\) is given by:
\[ \frac{a^2}{x_1}(x - x_1) + \frac{b^2}{y_1}(y - y_1) = a^2 + b^2 \]
Substituting \(x_1 = a\) and \(y_1 = \frac{b^2}{a}\):
\[ \frac{a^2}{a}(x - a) + \frac{b^2}{\frac{b^2}{a}}(y - \frac{b^2}{a}) = a^2 + b^2 \]
Simplifying gives:
\[ (x - a) + \frac{a}{b^2}(y - \frac{b^2}{a}) = a^2 + b^2 \]

4. **Finding the Intersection with the Axes**:
- **For point A (x-axis)**: Set \(y = 0\):
\[ (x - a) + \frac{a}{b^2}(0 - \frac{b^2}{a}) = a^2 + b^2 \]
This simplifies to:
\[ x - a - b = a^2 + b^2 \implies x = a + b + a^2 + b^2 \]
Thus, the coordinates of A are \((a + b + a^2 + b^2, 0)\).

- **For point B (y-axis)**: Set \(x = 0\):
\[ (0 - a) + \frac{a}{b^2}(y - \frac{b^2}{a}) = a^2 + b^2 \]
This simplifies to:
\[ -a + \frac{a}{b^2}y - 1 = a^2 + b^2 \implies \frac{a}{b^2}y = a + 1 + a^2 + b^2 \]
Thus, the coordinates of B are \((0, \frac{b^2}{a}(a + 1 + a^2 + b^2))\).

5. **Area of Triangle OAB**:
The area of triangle OAB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, base OA is \(x_A\) and height OB is \(y_B\):
\[ \text{Area} = \frac{1}{2} \times (a + b + a^2 + b^2) \times \frac{b^2}{a}(a + 1 + a^2 + b^2) \]

6. **Final Expression**:
After simplification, we find:
\[ \text{Area} = \frac{1}{2} \cdot \frac{b^2}{a} \cdot (a + b + a^2 + b^2)^2 \]