Hello there!

Greetings!


We know that ,Area of the quadrilateral inscribed in the ellipse i.e. kite

=1/2(product of its diagonals)



Now it is given, eccentricity e=3/5

Distance between foci= 6



Then,2ae=6

Or,ae=3

a(3/5)=3

Or, a=5

Now, b^2= a^2-(ae)^2
= b^2= 5^2-3^2
Or,b=4

Now, length of major axis, AA'=2a=10

And length of minor axis, BB'=2b=8

Required area

=Area of the kite ABA'B'

=1/2(AA')(BB')

=1/2(10)(8)

=1/2(80)

=40 square units


Thankyou