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Tushar Watts Grade: 12
        

Ques) From a fixed point P on the circumference of a circle of radius a , the perpendicualr PR is drawn to the tangent at Q (a variable point). Show that the max area of triangle PQR is [3 (3)1/2  / 8 ] a2.

8 years ago

Answers : (2)

dhwani kapoor
18 Points
										



make a line perpendicular to PQ and one perpendicular to QR and complete the rectangle.


 


join the centre O and point P. let the angle between OP and line perpendicular to PQ be x


OP = a


QR = asinx


PQ = a(cosx + 1)


 


area of PQR = asinx (cosx +1)/2


for maximizing,


differentiate


dA/dx = 2(cosx)^2 + cosx -1 = 0


cosx = pi (rejected) or pi/3


at pi/3 the next derivative is negative.


so it is a point of maxima. substituting the value, you get 3root3/8.

8 years ago
Ramesh V
70 Points
										

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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Regards,

Naga Ramesh

IIT Kgp - 2005 batch

8 years ago
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