Ellipse
Locus problems
Grade 11
Question:
<p>A stair-case of length \(l\) rests against a vertical wall and a floor of a room. Let \(P\) be a point on the stair-case, nearer to its end on the wall, that divides its length in the ratio \(1:2\). If the stair-case begins to slide on the floor, then the locus of \(P\) is</p>
<p>an ellipse of eccentricity \(\dfrac{1}{2}\).</p>
<p>an ellipse of eccentricity \(\dfrac{\sqrt{3}}{2}\).</p>
<p>a circle of radius \(\dfrac{l}{2}\).</p>
<p>a circle of radius \(\dfrac{\sqrt{3}}{2}\,l\).</p>
Step-by-Step Solution
Key Concept: Set up coordinates with the wall along y-axis and floor along x-axis. Use the constraint that the staircase length is constant (l) to derive the locus equation. The point P dividing the staircase in ratio 1:2 traces an elliptical path as the endpoints slide along the axes.
<p><strong>Step 1:</strong> Set up coordinates. Let the staircase touch the wall at A(0, a) and the floor at B(b, 0). The staircase length is constant: a² + b² = l²</p><p><strong>Step 2:</strong> Point P divides the staircase in ratio 1:2 (closer to wall). Using section formula, P divides AB in ratio 1:2 from A, so:<br>P = (2b/3, a/3)</p><p><strong>Step 3:</strong> Let P = (x, y). Then x = 2b/3 and y = a/3, giving b = 3x/2 and a = 3y</p><p><strong>Step 4:</strong> Substitute into the constraint a² + b² = l²:<br>(3y)² + (3x/2)² = l²<br>9y² + 9x²/4 = l²</p><p><strong>Step 5:</strong> Divide by l²:<br>x²/(4l²/9) + y²/(l²/9) = 1<br>x²/(2l/3)² + y²/(l/3)² = 1</p><p>This is an ellipse with semi-major axis 2l/3 and semi-minor axis l/3.</p><p>∴ Answer: B</p>
Correct Answer: B