Through the point `P(alpha,beta)` , where `alphabeta>0,` the straight line `x/a+y/b=1` is drawn so as to form a triangle of area `S` with the axes. If `a b >0,` then the least value of `S` is (a) `alphabeta` (b) `2alphabeta` (c) `3alphabeta` (d) none

Options

A. `alphabeta`
B. `2alphabeta`
C. `3alphabeta`
D. none

Explanation

To solve the problem, we need to find the least value of the area \( S \) of the triangle formed by the line \( \frac{x}{a} + \frac{y}{b} = 1 \) with the coordinate axes, given that the line passes through the point \( P(\alpha, \beta) \) where \( \alpha \beta > 0 \).

### Step-by-Step Solution:

1. **Identify the intercepts of the line**:
The line \( \frac{x}{a} + \frac{y}{b} = 1 \) intersects the x-axis when \( y = 0 \):
\[ \frac{x}{a} = 1 \implies x = a \quad \text{(x-intercept)} \]
It intersects the y-axis when \( x = 0 \):
\[ \frac{y}{b} = 1 \implies y = b \quad \text{(y-intercept)} \]

2. **Area of the triangle formed**:
The area \( S \) of the triangle formed by the line and the axes can be calculated using the formula for the area of a triangle:
\[ S = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times b \]

3. **Condition for the line to pass through point \( P(\alpha, \beta) \)**:
Since the line passes through \( P(\alpha, \beta) \), we can substitute these values into the line equation:
\[ \frac{\alpha}{a} + \frac{\beta}{b} = 1 \]

4. **Express \( a \) and \( b \) in terms of \( \alpha \) and \( \beta \)**:
Rearranging the equation gives:
\[ \frac{\alpha}{a} = 1 - \frac{\beta}{b} \implies a = \frac{\alpha b}{b - \beta} \]
Similarly, we can express \( b \) in terms of \( a \):
\[ \frac{\beta}{b} = 1 - \frac{\alpha}{a} \implies b = \frac{\beta a}{a - \alpha} \]

5. **Substitute \( a \) and \( b \) into the area formula**:
Substitute \( a \) and \( b \) into the area formula:
\[ S = \frac{1}{2} \times \frac{\alpha b}{b - \beta} \times b = \frac{\alpha b^2}{2(b - \beta)} \]
Similarly, we can substitute for \( b \) to express \( S \) in terms of \( a \).

6. **Minimize the area \( S \)**:
To find the minimum area, we can use the method of Lagrange multipliers or analyze the function directly. However, we can also use the AM-GM inequality:
\[ S = \frac{1}{2} \times a \times b \geq \frac{1}{2} \times 2\sqrt{ab} \times 2\sqrt{ab} = 2\alpha\beta \]
The minimum area occurs when \( a = b \), leading to:
\[ S \geq 2\alpha\beta \]

### Conclusion:
Thus, the least value of the area \( S \) is:
\[ \boxed{2\alpha\beta} \]