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…
JEE Main 2014 — Mathematics Coordinate Geometry
2014mcqmedium
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 staircase begins to slide on the floor, then the locus of P is:
Official previous-year question
Held on 11 Apr 2014 · Verified 6 Jul 2026.
Options
A
an ellipse of eccentricity 21
B
an ellipse of eccentricity 23
C
a circle of radius 21
D
a circle of radius 23l
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Solution
Let point A(a,0) is on x-axis and B(0,b) is on y-axis.
Let P(h,k) divides AB in the ratio 1:2. So, by section formula h=1+22(0)+1(a)=3ak=32(b)+1(0)=32b⇒a=3h and b=23k Now, a2+b2=l2⇒9h2+49k2=l2⇒(3l)2h2+(32l)2k2=1 Now e=1−(9l2×4l29)=1−41=23 Thus, required locus of P is an ellipse with eccentricity 23.
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