Ramesh V
Last Activity: 15 Years ago
let the circle be x2 + y2 =a2
here a is radius of circle
let fixed point be P(a,0)
variable point Q(a cos x , a sin x )
so R (a , a sin x )
so, area of triangle PQR is
area = 1/2* | a a acosx a | | 0 asinx asinx 0 |
area = a2/2*( sinx - sinx.cosx)
to find max area d(area)/dx = 0
which gives cosx = -1/2 so is sinx = 31/2 /2
so max. area comes to be : 3*31/2*a2 /8
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Naga Ramesh
IIT Kgp - 2005 batch