1. Introduction to Key Concepts and Formulas

To master this problem, we'll seamlessly integrate several fundamental concepts from coordinate geometry and basic calculus (for areas). Here's a quick refresher on the tools we'll be using:

• Area of a Rectangle: For a rectangle with length $L$ and width $W$, its area is simply $A = L \times W$. We'll use this to find the total area of rectangle R.
• Area of a Right-Angled Triangle: Given a right-angled triangle with base $b$ and height $h$, its area is $A = \frac{1}{2} \times b \times h$. This will be crucial for calculating the area of the triangular region formed by the line segment AB.
• Midpoint Formula: If we have two points, $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the coordinates of their midpoint $M(h,k)$ are given by the average of their respective coordinates: $h = \frac{x_1+x_2}{2} \quad \text{and} \quad k = \frac{y_1+y_2}{2}$ This formula is our bridge to connect the points A and B to their midpoint.
• Locus of a Point: The locus of a point is the path or curve traced by the point as it moves according to specific conditions. To find the equation of a locus, we typically assign generic coordinates $(h,k)$ to the moving point, establish an algebraic relationship between $h$ and $k$ based on the problem's conditions, and then replace $h$ with $x$ and $k$ with $y$ to get the final equation in standard Cartesian form.
• Equation of a Hyperbola: A hyperbola is one of the conic sections. A particularly relevant form for this problem is the rectangular hyperbola (also known as an equilateral hyperbola), whose standard equation is often given as $xy=c^2$ (or $xy=k$), where its asymptotes are the coordinate axes.

2. Visualizing the Rectangle and Calculating its Total Area

Our first step is to clearly define the boundaries of the rectangle R and calculate its total area. This provides the context for the area division.

The rectangle R is defined by the following lines:

• $x=0$ (the y-axis)
• $x=2$ (a vertical line parallel to the y-axis)
• $y=0$ (the x-axis)
• $y=5$ (a horizontal line parallel to the x-axis)

Why this step? Visualizing the rectangle helps us understand the geometry of the problem. Knowing the total area is essential because the line segment AB divides this total area into a given ratio.

Let's identify the vertices of this rectangle: $P(0,0)$, $Q(2,0)$, $S(0,5)$, and $T(2,5)$.

• The length of the rectangle (along the x-axis, from $x=0$ to $x=2$) is $L = 2 - 0 = 2$ units.
• The width of the rectangle (along the y-axis, from $y=0$ to $y=5$) is $W = 5 - 0 = 5$ units.

The total area of rectangle R, which we'll denote as $A_R$, is: $A_R = L \times W = 2 \times 5 = 10 \text{ square units}$

3. Analyzing the Area Division by Line Segment AB

Next, we introduce the points A and B and understand how the line segment AB divides the rectangle's area.

We are given two points:

• $A(\alpha,0)$: This point lies on the x-axis ($y=0$). The condition $\alpha \in [0,2]$ means it is on the bottom edge of the rectangle.
• $B(0,\beta)$: This point lies on the y-axis ($x=0$). The condition $\beta \in [0,5]$ means it is on the left edge of the rectangle.

The line segment AB connects these two points. Observe that together with the origin $P(0,0)$, the point $A(\alpha,0)$, and the point $B(0,\beta)$, these form a right-angled triangle $\triangle PAB$ at the origin.

The problem states that the line segment AB divides the total area of the rectangle R in the ratio $4:1$. This means the total area ($10$ square units) is split into two parts:

• One part is $\frac{4}{4+1} = \frac{4}{5}$ of the total area.
• The other part is $\frac{1}{4+1} = \frac{1}{5}$ of the total area.

Why this step? It's critical to correctly identify which region corresponds to which ratio. By visualizing (or sketching) the rectangle and the line segment AB, it's clear that the triangle $\triangle PAB$ (in the bottom-left corner) is a smaller region compared to the remaining pentagonal area. Therefore, the area of $\triangle PAB$ must correspond to the smaller ratio.

So, the area of $\triangle PAB$ is $\frac{1}{5}$ of the total area of rectangle R: $\text{Area}(\triangle PAB) = \frac{1}{5} \times A_R = \frac{1}{5} \times 10 = 2 \text{ square units}$

Now, let's calculate the area of $\triangle PAB$ using the coordinates of A and B.

• The base of $\triangle PAB$ (along the x-axis) is the distance from $P(0,0)$ to $A(\alpha,0)$, which is $\alpha$.
• The height of $\triangle PAB$ (along the y-axis) is the distance from $P(0,0)$ to $B(0,\beta)$, which is $\beta$.

Using the formula for the area of a right-angled triangle: $\text{Area}(\triangle PAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \alpha \times \beta$

Equating the two expressions for the area of $\triangle PAB$: $\frac{1}{2} \alpha \beta = 2$ Multiplying both sides by 2, we obtain a crucial relationship between $\alpha$ and $\beta$: $\alpha \beta = 4 \quad \ldots(1)$

Constraint Check: Let's quickly verify if this relationship is consistent with the given ranges $\alpha \in [0,2]$ and $\beta \in [0,5]$. Since $\alpha\beta=4$, neither $\alpha$ nor $\beta$ can be zero.

• If $\alpha$ takes its maximum value of $2$, then $\beta = 4/2 = 2$. This value of $\beta$ is within its allowed range $[0,5]$.
• If $\beta$ takes its maximum value of $5$, then $\alpha = 4/5$. This value of $\alpha$ is within its allowed range $[0,2]$. These checks confirm that such points A and B can indeed exist within the boundaries of the rectangle.

4. Finding the Locus of the Midpoint of AB

Our ultimate goal is to find the path traced by the midpoint of AB. Let's denote the coordinates of this midpoint as $M(h,k)$.

We use the midpoint formula for points $A(\alpha,0)$ and $B(0,\beta)$:

• The x-coordinate of the midpoint is $h = \frac{\alpha + 0}{2} = \frac{\alpha}{2}$.
• The y-coordinate of the midpoint is $k = \frac{0 + \beta}{2} = \frac{\beta}{2}$.

Why this step? The midpoint formula directly relates the coordinates of the moving point $(h,k)$ to the variables $\alpha$ and $\beta$. Our strategy is to eliminate $\alpha$ and $\beta$ to find an equation solely in terms of $h$ and $k$.

From these equations, we can express $\alpha$ and $\beta$ in terms of $h$ and $k$:

• $\alpha = 2h$
• $\beta = 2k$

Now, substitute these expressions for $\alpha$ and $\beta$ into the fundamental relationship we derived in Equation (1), which is $\alpha \beta = 4$: $(2h)(2k) = 4$ $4hk = 4$ Dividing both sides by 4: $hk = 1$

Why this step? This substitution is the core of finding the locus. By replacing the auxiliary variables $\alpha$ and $\beta$ with expressions involving $h$ and $k$, we directly obtain an equation that describes the path of the midpoint $M$.

Finally, to express the locus in standard Cartesian form, we replace $h$ with $x$ and $k$ with $y$: $xy = 1$

5. Identifying the Curve

The equation $xy=1$ is a standard form of a rectangular hyperbola.

Why this identification? Recognizing the form of the equation is the final step in classifying the locus. A rectangular hyperbola is characterized by its asymptotes being perpendicular, in this case, the x-axis ($y=0$) and the y-axis ($x=0$).

Therefore, the midpoint of AB lies on a hyperbola.

Tips for Success and Common Mistakes to Avoid:

• Visualize the Problem: Always begin by sketching the rectangle, the points A and B, and the line segment AB. A clear diagram prevents misinterpretations of the area division.
• Correct Area Ratio Interpretation: When an area is divided in the ratio $m:n$, the individual parts are $\frac{m}{m+n}$ and $\frac{n}{m+n}$ of the total area. In this problem, $4:1$ means $1/5$ and $4/5$. Make sure you correctly assign the smaller area (the triangle) to the smaller ratio ($1/5$).
• Algebraic Precision: Pay close attention to your substitutions. Ensure $\alpha$ and $\beta$ are correctly expressed in terms of $h$ and $k$ before substituting them into the $\alpha\beta$ relationship. Small errors here can lead to incorrect locus equations.
• Locus Convention: Remember the final step of replacing $(h,k)$ with $(x,y)$ to present the locus equation in its standard form.
• Constraint Awareness: While not directly affecting the type of curve here, always be mindful of variable constraints (e.g., $\alpha \in [0,2]$, $\beta \in [0,5]$). They ensure the problem is well-defined and that the points exist within the specified region.

Summary and Key Takeaway:

This problem elegantly combines geometric understanding with algebraic manipulation to determine a locus. We started by calculating the total area of the rectangle. By carefully interpreting the given area ratio, we deduced the area of the right-angled triangle formed by the line segment AB and the coordinate axes. This allowed us to establish a crucial algebraic relationship between the coordinates $\alpha$ and $\beta$ of points A and B, specifically $\alpha\beta=4$. We then used the midpoint formula to express $\alpha$ and $\beta$ in terms of the midpoint's coordinates $(h,k)$. Substituting these expressions back into the $\alpha\beta$ relationship yielded an equation solely in terms of $h$ and $k$, which was $hk=1$. Finally, by replacing $(h,k)$ with $(x,y)$, we identified the locus as the equation of a hyperbola, $xy=1$. This problem is a great illustration of how to systematically approach locus problems by building connections between geometric properties and algebraic equations.