If the roots of `a_1x^(2)+b_1x+c_1=0 and alpha_1 , beta_1` and those of ` a_2x^(2)+b_2x+c_2=0` are ` alpha_2 , beta_2` such that `alpha_1alpha_2=beta_1beta_2=1` then :

Options

A. `(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)`
B. `(a_1)/(c_2)=(b_1)/(b_2)=(c_1)/(a_2)`
C. `a_1a_2=b_1b_2=c_1c_2`
D. None of these

Answer

To solve the problem, we need to analyze the relationships between the coefficients of the two quadratic equations based on the given conditions about their roots.

### Step-by-Step Solution:

1. **Identify the equations and roots**:
We have two quadratic equations:
- \( a_1x^2 + b_1x + c_1 = 0 \) with roots \( \alpha_1, \beta_1 \)
- \( a_2x^2 + b_2x + c_2 = 0 \) with roots \( \alpha_2, \beta_2 \)

2. **Use Vieta's formulas**:
According to Vieta's formulas:
- For the first equation:
- Sum of roots: \( \alpha_1 + \beta_1 = -\frac{b_1}{a_1} \)
- Product of roots: \( \alpha_1 \beta_1 = \frac{c_1}{a_1} \)
- For the second equation:
- Sum of roots: \( \alpha_2 + \beta_2 = -\frac{b_2}{a_2} \)
- Product of roots: \( \alpha_2 \beta_2 = \frac{c_2}{a_2} \)

3. **Apply the given conditions**:
We are given that:
- \( \alpha_1 \alpha_2 = 1 \)
- \( \beta_1 \beta_2 = 1 \)

From these, we can express \( \alpha_1 \) and \( \beta_1 \) in terms of \( \alpha_2 \) and \( \beta_2 \):
- \( \alpha_1 = \frac{1}{\alpha_2} \)
- \( \beta_1 = \frac{1}{\beta_2} \)

4. **Substitute into the sum of roots**:
Substitute \( \alpha_1 \) and \( \beta_1 \) into the sum of roots equation:
\[ \frac{1}{\alpha_2} + \frac{1}{\beta_2} = -\frac{b_1}{a_1} \]
This can be rewritten as:
\[ \frac{\beta_2 + \alpha_2}{\alpha_2 \beta_2} = -\frac{b_1}{a_1} \]

5. **Substitute the sum and product of roots**:
Substitute \( \alpha_2 + \beta_2 \) and \( \alpha_2 \beta_2 \) using Vieta's formulas:
\[ \frac{-\frac{b_2}{a_2}}{\frac{c_2}{a_2}} = -\frac{b_1}{a_1} \]
Simplifying gives:
\[ \frac{b_2}{c_2} = \frac{b_1}{a_1} \]

6. **Repeat for product of roots**:
Substitute \( \alpha_1 \) and \( \beta_1 \) into the product of roots equation:
\[ \frac{1}{\alpha_2} \cdot \frac{1}{\beta_2} = \frac{c_1}{a_1} \]
This can be rewritten as:
\[ \frac{1}{\alpha_2 \beta_2} = \frac{c_1}{a_1} \]
Substitute \( \alpha_2 \beta_2 = \frac{c_2}{a_2} \):
\[ \frac{a_2}{c_2} = \frac{c_1}{a_1} \]

7. **Combine the results**:
From the two results, we have:
\[ \frac{b_2}{b_1} = \frac{c_2}{a_1} = \frac{a_2}{c_1} \]
This gives us the final relationship as:
\[ \frac{b_2}{b_1} = \frac{c_2}{a_1} = \frac{a_2}{c_1} \]

### Final Answer:
Thus, the correct relation is:
\[ \frac{b_2}{b_1} = \frac{c_2}{a_1} = \frac{a_2}{c_1} \]