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Question

Rectangle ABCD has area 200. An ellipse with area 200π passes through A and C and has foci at B and D. If the perimeter of the rectangle is P, then the value of P20 is

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Solution

Let the sides of rectangle be p and q.
Area of rectangle =pq=200(i)

Area of ellipse =πab=200π
ab=200(ii)
Perimeter of rectangle =2(p+q)
From triangle ABD,
BD= Distance between foci
p2+q2=4a2e2
(p+q)22pq=4(a2b2)(iii)
Also, from the definition of ellipse, the sum of focal lengths is 2a. Then,
AB+AD=p+q=2a(iv)
Putting the value of (p+q) in (iii),
(2a)22pq=4a24b2
4a22×200=4(a2b2)
a2100=a2b2
b=10
From (ii),
ab=200a=20
We know that,
p+q=2a
So the perimeter will be
2(p+q)=4×20=80
Hence P20=4

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