Let `A_(r)` be the area of the region bounded between the curves `y^(2)=(e^(-kr))x("where "k gt0,r in N)" and the line "y=mx ("where "m ne 0)`, k and m are some constants
`A_(1),A_(2),A_(3),…` are in G.P. with common ratio

Options

A. `e^(-k)`
B. `e^(-2k)`
C. `e^(-4k)`
D. None of these

Explanation

To solve the problem, we need to find the area \( A_r \) bounded by the curve \( y^2 = e^{-kr} x \) and the line \( y = mx \), where \( k > 0 \) and \( r \in \mathbb{N} \). We will also show that the areas \( A_1, A_2, A_3, \ldots \) are in geometric progression (G.P.) and find the common ratio.

### Step 1: Find the Intersection Points
We start by finding the intersection points of the curve and the line. We have:
1. The curve: \( y^2 = e^{-kr} x \)
2. The line: \( y = mx \)

Substituting \( y = mx \) into the curve's equation gives:
\[ (mx)^2 = e^{-kr} x \]
This simplifies to:
\[ m^2 x^2 = e^{-kr} x \]
Rearranging, we get:
\[ m^2 x^2 - e^{-kr} x = 0 \]
Factoring out \( x \):
\[ x(m^2 x - e^{-kr}) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( x = \frac{e^{-kr}}{m^2} \)

### Step 2: Set Up the Integral for Area
The area \( A_r \) between the curves from \( x = 0 \) to \( x = \frac{e^{-kr}}{m^2} \) can be calculated using the integral:
\[ A_r = \int_0^{\frac{e^{-kr}}{m^2}} (y_{\text{curve}} - y_{\text{line}}) \, dx \]
Where:
- \( y_{\text{curve}} = \sqrt{e^{-kr} x} \)
- \( y_{\text{line}} = mx \)

Thus, the area becomes:
\[ A_r = \int_0^{\frac{e^{-kr}}{m^2}} \left( \sqrt{e^{-kr} x} - mx \right) \, dx \]

### Step 3: Calculate the Integral
Now we compute the integral:
1. **Integral of \( \sqrt{e^{-kr} x} \)**:
\[ \int \sqrt{e^{-kr} x} \, dx = \sqrt{e^{-kr}} \int \sqrt{x} \, dx = \sqrt{e^{-kr}} \cdot \frac{2}{3} x^{3/2} \]
Evaluating from \( 0 \) to \( \frac{e^{-kr}}{m^2} \):
\[ = \sqrt{e^{-kr}} \cdot \frac{2}{3} \left( \frac{e^{-kr}}{m^2} \right)^{3/2} = \frac{2}{3} \cdot \frac{e^{-kr/2}}{m^3} \]

2. **Integral of \( mx \)**:
\[ \int mx \, dx = \frac{m}{2} x^2 \]
Evaluating from \( 0 \) to \( \frac{e^{-kr}}{m^2} \):
\[ = \frac{m}{2} \left( \frac{e^{-kr}}{m^2} \right)^2 = \frac{m e^{-2kr}}{2m^4} = \frac{e^{-2kr}}{2m^3} \]

### Step 4: Combine the Results
Now we can combine the results of the integrals:
\[ A_r = \frac{2}{3} \cdot \frac{e^{-kr/2}}{m^3} - \frac{e^{-2kr}}{2m^3} \]
Factoring out \( \frac{e^{-2kr}}{m^3} \):
\[ A_r = \frac{e^{-2kr}}{m^3} \left( \frac{2}{3} e^{kr/2} - \frac{1}{2} \right) \]

### Step 5: Check for G.P. Condition
Now we will find \( A_1, A_2, A_3 \) for \( r = 1, 2, 3 \):
- \( A_1 = \frac{e^{-2k}}{m^3} \left( \frac{2}{3} e^{k/2} - \frac{1}{2} \right) \)
- \( A_2 = \frac{e^{-4k}}{m^3} \left( \frac{2}{3} e^{k} - \frac{1}{2} \right) \)
- \( A_3 = \frac{e^{-6k}}{m^3} \left( \frac{2}{3} e^{3k/2} - \frac{1}{2} \right) \)

To show they are in G.P., we need to check:
\[ \frac{A_2}{A_1} = \frac{A_3}{A_2} \]
Calculating \( \frac{A_2}{A_1} \):
\[ \frac{A_2}{A_1} = \frac{e^{-4k}}{e^{-2k}} \cdot \frac{\frac{2}{3} e^{k} - \frac{1}{2}}{\frac{2}{3} e^{k/2} - \frac{1}{2}} = e^{-2k} \cdot \text{(some expression)} \]
Calculating \( \frac{A_3}{A_2} \):
\[ \frac{A_3}{A_2} = \frac{e^{-6k}}{e^{-4k}} \cdot \frac{\frac{2}{3} e^{3k/2} - \frac{1}{2}}{\frac{2}{3} e^{k} - \frac{1}{2}} = e^{-2k} \cdot \text{(same expression)} \]

Since both ratios are equal, \( A_1, A_2, A_3 \) are in G.P. with common ratio \( e^{-2k} \).

### Final Answer
The common ratio of the areas \( A_1, A_2, A_3 \) is \( e^{-2k} \).