Question

A rectangle is drawn by lines x=0, x=2, y=0 and y=5. Points A and B lie on coordinate axes. If line AB divides the area of rectangle in 4:1, then the locus of mid-point of AB is?

• A. Circle
• B. Hyperbola
• C. Ellipse
• D. Straight line

Answer 1

The Correct Option is B

To find the locus of the midpoint of line segment AB, where A and B lie on the coordinate axes, we begin with understanding the geometry and condition given.

1. The rectangle is bounded by the lines: \(x = 0, x = 2, y = 0,\) and \(y = 5.\) Thus, the rectangle's dimensions are 2 units (width) and 5 units (height), covering an area of \(2 \times 5 = 10\) square units.
2. Points A and B are on the coordinate axes, hence let \(A = (a, 0)\) on the x-axis and \(B = (0, b)\) on the y-axis.
3. The equation for line AB can be represented using the intercept form: \(\frac{x}{a} + \frac{y}{b} = 1.\)
4. The problem states the line divides the rectangle into areas of ratio 4:1. This means one section (say section above the line) has an area of \(8\) units, and below the line, it's \(2\) units.
5. To find the midpoint of AB, we use the midpoint formula: \(M = \left(\frac{a}{2}, \frac{b}{2}\right).\)
6. Since the line divides the rectangle into a specific area ratio, it satisfies the condition by being a constant product of \(2ab = k\) (for some constant \(k\)).
7. In terms of the midpoint, let \(x = \frac{a}{2}\) and \(y = \frac{b}{2}.\) This gives the relationship:

\(2 \times (2x)(2y) = k \implies 8xy = k.\)

Simplifying, we find:

\(xy = \frac{k}{8},\) which is the standard form of a hyperbola.

1. This confirms that the locus of the midpoint of AB is a hyperbola. Thus:

The correct answer is Hyperbola.