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Question:

A staircase of length l rests against a vertical wall and a floor of a room. Let P be a point on the staircase, nearer to its end on the wall, that divides its length in the ratio 1:2. If the staircase begins to slide on the floor, then the locus of P is:

An ellipse of eccentricity 1/2

An ellipse of eccentricity √3/2

A circle of radius l/2

A circle of radius √3/2l

Solution:

Let b be the height and a be the length intercepted by the staircase.
By section formula, we can write the coordinates of P as: (b/3, 2a/3)
Now, the length of the staircase is constant.
Hence, a² + b² = l²
Let P be (x, y). Hence, (3x)² + (3y/2)² = l²
=> x²/l²/9 + y²/4l²/9 = 1
This represents the equation of an ellipse, 1 - e² = 1/4
Hence, e = √3/2