Parabola

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

3 Answers

11
Joydoot ghatak ·

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...

1
Vinay Arya ·

Thank you very much.

6
AKHIL ·

it can also be done by maxima and minima...

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