The asymptotes of the hyperbola `(x^(2))/(a_(1)^(2))-(y^(2))/(b_(1)^(2))=1` and `(x^(2))/(a_(2)^(2))-(y^(2))/(b_(2)^(2))=1` are perpendicular to each other. Then, (a) `a_(1)/a_(2)=b_(1)/b_(2)` (b) `a_(1)a_(2)=b_(1)b_(2)` (c) `a_(1)a_(2)+b_(1)b_(2)=0` (d) `a_(1)-a_(2)=b_(1)-b_(2)`

Options

A. `a_(1)//a_(2)=b_(1)//b_(2)`
B. `a_(1)a_(2)=b_(1)b_(2)`
C. `a_(1)a_(2)+b_(1)b_(2)=0`
D. `a_(1)-a_(2)=b_(1)-b_(2)`

Explanation

To solve the problem, we need to analyze the asymptotes of the given hyperbolas and establish the relationship between their parameters when the asymptotes are perpendicular.

### Step-by-Step Solution:

1. **Identify the equations of the hyperbolas**:
The equations of the hyperbolas are given as:
\[ \frac{x^2}{a_1^2} - \frac{y^2}{b_1^2} = 1 \quad \text{(1)} \]
\[ \frac{x^2}{a_2^2} - \frac{y^2}{b_2^2} = 1 \quad \text{(2)} \]

2. **Determine the asymptotes of each hyperbola**:
The asymptotes for the hyperbola in standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by:
\[ y = \pm \frac{b}{a} x \]
For the first hyperbola (1), the asymptotes are:
\[ y = \pm \frac{b_1}{a_1} x \quad \text{(3)} \]
For the second hyperbola (2), the asymptotes are:
\[ y = \pm \frac{b_2}{a_2} x \quad \text{(4)} \]

3. **Identify the slopes of the asymptotes**:
From equations (3) and (4), we can identify the slopes:
- Slope of the first hyperbola, \(m_1 = \frac{b_1}{a_1}\)
- Slope of the second hyperbola, \(m_2 = \frac{b_2}{a_2}\)

4. **Condition for perpendicularity**:
The asymptotes are perpendicular if the product of their slopes is \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \frac{b_1}{a_1} \cdot \frac{b_2}{a_2} = -1 \]

5. **Rearranging the equation**:
This leads to:
\[ b_1 b_2 = -a_1 a_2 \]

6. **Final expression**:
We can rearrange the above equation to express it in a more useful form:
\[ a_1 a_2 + b_1 b_2 = 0 \]

7. **Conclusion**:
Therefore, the correct option that represents the relationship between \(a_1, a_2, b_1, b_2\) is:
\[ \text{(c) } a_1 a_2 + b_1 b_2 = 0 \]