sum1 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.
-
UP 0 DOWN 0 0 8
8 Answers
iitimcomin
·2009-06-24 08:58:10
(3y/l)^2 + (3x/2l)^2 = 1 [takin point of intersection of those two[perp] lines as origin]
Manish Shankar
·2009-06-24 09:05:00
take 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
Mirka
·2009-06-24 22:04:32
locus of the reqd point will be
(3x/2)2 + (3y)2 = L2
or
(3y/2)2 + (3x)2 = L2
It's an ellipse.