Let R be a rectangle given by the lines $$x = 0$$, $$x = 2$$, $$y = 0$$ and $$y = 5$$. Let $$A(\alpha, 0)$$ and $$B(0, \beta)$$, $$\alpha \in (0, 2)$$ and $$\beta \in (0, 5)$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4:1. Then, the mid-point of AB lies on a

We need to find the locus of the midpoint of segment AB, where $$A(\alpha, 0)$$ and $$B(0, \beta)$$ are points on the sides of rectangle R such that AB divides the rectangle's area in ratio 4:1.

To begin,

The rectangle R has vertices at $$(0,0)$$, $$(2,0)$$, $$(2,5)$$, and $$(0,5)$$.

Area of R = $$2 \times 5 = 10$$ square units.

Next,

The line segment AB connects $$A(\alpha, 0)$$ on the x-axis to $$B(0, \beta)$$ on the y-axis. Together with the origin $$O(0,0)$$, this forms a right triangle OAB with legs along the axes.

$$ \text{Area of } \triangle OAB = \frac{1}{2} \times \alpha \times \beta $$

From here,

The line segment AB divides the rectangle into two parts in the ratio 4:1. Since the total area is 10, the two parts have areas 8 and 2. The triangle OAB (in the corner) is the smaller part, so:

$$ \frac{1}{2}\alpha\beta = 2 $$

$$ \alpha\beta = 4 \quad \cdots (1) $$

Continuing,

Let the midpoint $$M$$ of AB have coordinates $$(h, k)$$:

$$ h = \frac{\alpha + 0}{2} = \frac{\alpha}{2}, \qquad k = \frac{0 + \beta}{2} = \frac{\beta}{2} $$

Therefore: $$\alpha = 2h$$ and $$\beta = 2k$$.

Now,

Substituting $$\alpha = 2h$$ and $$\beta = 2k$$ into equation (1):

$$ (2h)(2k) = 4 $$

$$ 4hk = 4 $$

$$ hk = 1 $$

Replacing $$(h, k)$$ with $$(x, y)$$, the locus is:

$$ xy = 1 $$

This is the equation of a rectangular hyperbola.

The correct answer is Option 3: hyperbola.