`alpha_(1),beta_(1)` are the roots of `ax^(2)+bx+c=0 and alpha_(2), beta_(2)` are the roots of `px^(2) +qx+r=0` If `alpha_(1) alpha_(2)=beta_(1) beta_(2)=1` then

Options

A. `(a)/(p)=(b)/(q)=(c )/(r )`
B. `(a)/(r )=(b)/(q)=(c )/(p)`
C. `ap=bq=cr`
D. none

Answer

To solve the problem step by step, we will analyze the given quadratic equations and the relationships between their roots.

### Step 1: Identify the roots and their relationships
We have two quadratic equations:
1. \( ax^2 + bx + c = 0 \) with roots \( \alpha_1 \) and \( \beta_1 \)
2. \( px^2 + qx + r = 0 \) with roots \( \alpha_2 \) and \( \beta_2 \)

Given that:
- \( \alpha_1 \alpha_2 = 1 \)
- \( \beta_1 \beta_2 = 1 \)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- For the first equation:
- Sum of roots: \( \alpha_1 + \beta_1 = -\frac{b}{a} \) (Equation 1)
- Product of roots: \( \alpha_1 \beta_1 = \frac{c}{a} \) (Equation 2)

- For the second equation:
- Sum of roots: \( \alpha_2 + \beta_2 = -\frac{q}{p} \) (Equation 3)
- Product of roots: \( \alpha_2 \beta_2 = \frac{r}{p} \) (Equation 4)

### Step 3: Substitute the product of roots
From the given conditions:
- Since \( \alpha_1 \alpha_2 = 1 \), we can express \( \alpha_2 \) in terms of \( \alpha_1 \):
\[ \alpha_2 = \frac{1}{\alpha_1} \]
- Similarly, since \( \beta_1 \beta_2 = 1 \):
\[ \beta_2 = \frac{1}{\beta_1} \]

### Step 4: Substitute into Vieta's formulas
Substituting \( \alpha_2 \) and \( \beta_2 \) into the product of roots for the second equation:
\[ \alpha_2 \beta_2 = \frac{1}{\alpha_1} \cdot \frac{1}{\beta_1} = \frac{1}{\alpha_1 \beta_1} \]
Setting this equal to the product of roots from Vieta's:
\[ \frac{1}{\alpha_1 \beta_1} = \frac{r}{p} \]
From Equation 2, we know \( \alpha_1 \beta_1 = \frac{c}{a} \):
\[ \frac{1}{\frac{c}{a}} = \frac{r}{p} \implies \frac{a}{c} = \frac{r}{p} \]
This gives us:
\[ ar = cp \quad \text{(Equation 5)} \]

### Step 5: Analyze the sum of roots
Now, substituting \( \alpha_2 \) and \( \beta_2 \) into the sum of roots for the second equation:
\[ \alpha_2 + \beta_2 = \frac{1}{\alpha_1} + \frac{1}{\beta_1} = \frac{\beta_1 + \alpha_1}{\alpha_1 \beta_1} \]
Using Equation 1:
\[ \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \]
Setting this equal to the sum of roots from Vieta's for the second equation:
\[ -\frac{q}{p} = -\frac{b}{c} \implies \frac{q}{p} = \frac{b}{c} \quad \text{(Equation 6)} \]

### Step 6: Final relationships
From Equations 5 and 6, we have:
1. \( ar = cp \)
2. \( q = \frac{b}{c}p \)

### Conclusion
The relationships derived from the roots lead us to the conclusion that:
- \( \frac{a}{r} = \frac{b}{q} = \frac{c}{p} \)