1. Rectangle and intercept form of the line

The rectangle is bounded by: $x=0,\quad x=2,\quad y=0,\quad y=5$ so its area is $\text{Area}(R)=2\times 5=10.$

Given points: $A(\alpha,0),\qquad B(0,\beta),$ with $\alpha\in[0,2],\qquad \beta\in[0,5].$

The segment $AB$ together with the coordinate axes forms a right triangle $OAB$ where $O=(0,0)$. Its area is $\text{Area}(\triangle OAB)=\frac12\alpha\beta.$

2. Using the area ratio condition

The segment $AB$ divides the rectangle into two parts in the ratio $4:1$. Since the total area is $10$, the two parts must have areas $8 \text{ and } 2.$

Inside the rectangle, the smaller region is the triangle near the origin bounded by the axes and $AB$, so $\frac12\alpha\beta=2.$ Hence, $\alpha\beta=4.$

(If we had taken the triangle area as $8$, then $\alpha\beta=16$, impossible since $\alpha\le 2$ and $\beta\le 5$ imply $\alpha\beta\le 10$.)

3. Coordinates of the midpoint of }AB

Let the midpoint of $AB$ be $M(h,k)$. Then $M\left(\frac\alpha2,\frac\beta2\right).$ So, $h=\frac\alpha2,\qquad k=\frac\beta2.$ Thus, $\alpha=2h,\qquad \beta=2k.$

Substitute into $\alpha\beta=4$: $(2h)(2k)=4$ $4hk=4$ $hk=1.$

Therefore the locus of the midpoint is $xy=1,$ which is a rectangular hyperbola.

4. Checking options

• A: hyperbola — Correct, since the locus is $xy=1$.
• B: straight line — Incorrect.
• C: parabola — Incorrect.
• D: circle — Incorrect.

Hence, the midpoint of $AB$ lies on a hyperbola.