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# 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.

## 2 Answers

11 years ago

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.

11 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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