Normal drawn at P(t) having negative ordinate,to parabola y2=10x meets the curve again at Q(t2).If t2 is least then length PQ is
-
UP 0 DOWN 0 0 3
3 Answers
Joydoot ghatak
·2011-03-16 22:36:28
for a parabola,
if normal at t cuts the parabola again at t2, then
t2 = -t -2t. (this can be proved..)
as t is -ve.
-t and -2/t both are +ve..
thus we can apply A.M. ≥ G.M.
thus, (-t)+(-2/t)2 ≥ √(-t).(-2/t)
thus, t2 ≥ 2√2
least value of t2 = 2√2.
thus, t= -√2.
y2 =4 .52 x
thus, a= 5/2
thus P = (at2,2at) = (5, -5√2)
thus, Q = (at22,2at2) = (20, 10√2)
thus, PQ = 15√3.
i am getting this as answer...