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Let Q = (3,sqrt(5)),R =(7,3sqrt(5)). A p...

Let `Q = (3,sqrt(5)),R =(7,3sqrt(5))`. A point P in the XY-plane varies in such a way that perimeter of `DeltaPQR` is 16. Then the maximum area of `DeltaPQR` is

A

6

B

12

C

18

D

9

Text Solution

Verified by Experts

The correct Answer is:
B
P lies on the ellipse for which Q and R are foci such that `QR = 2ae =6` and `2a =` perimeter of triangle `-2ae = 16 - 6 = 10`
`:. a = 5` and `e = 3//5`, also `b^(2) = a^(2) -a^(2) e^(2) = 25 -9 = 16`
`b = 4`
Maximum area of triangle `=(1)/(2) (2ae)(b) = 12` sq. units
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