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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.`
10 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.
```
10 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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Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
10 years ago
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