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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 78

A stair-case of length ll 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 :

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Answer and worked solution

Official answer: B

Put the foot on the floor at (lcos⁡θ,0)(l\cos\theta,0) and the top at (0,lsin⁡θ)(0,l\sin\theta). P is l3\frac l3 from the top, so P=(l3cos⁡θ,2l3sin⁡θ)P=\left(\frac l3\cos\theta,\frac{2l}3\sin\theta\right).

This is an ellipse with semi-axes l3\frac l3 and 2l3\frac{2l}3, so e=1−14=32e=\sqrt{1-\frac14}=\dfrac{\sqrt3}2.

Topics:LocusEllipse
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