 # 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.

14 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.

14 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

----

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch