11sum1 pls do this
The ends of a rod of length l moves on 2 mutually perpendicular lines.Find the locus of the point on the rod which divides it in the ratio 2:1.
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8 Answers
3(3y/l)^2 + (3x/2l)^2 = 1 [takin point of intersection of those two[perp] lines as origin]
1357take the ends of rods as (x,0) and (0,y)
required point P is find the point (h,k) on the rod which divides the length in ratio 2:1
the put h, k in x2+y2=l2
1locus of the reqd point will be
(3x/2)2 + (3y)2 = L2
or
(3y/2)2 + (3x)2 = L2
It's an ellipse.