Question

If the area of the ellipse `((x^2)/(a^2))+((y^2)/(b^2))=1` is `4pi` , then find the maximum area of rectangle inscribed in the ellipse.

Answer 1

Correct Answer - 8 sq. units
The area of ellipse `(x^(2)//a^(2))+(y^(2)//b^(2))=1 "is" piab=4pi` (given) . Therefore, ab=4
`A(a cos theta, b sin theta)`
on the ellipse.
Now, the area of reactangle ABCD inscribed in the ellipse is

`Delta=4(a cos theta)(b sin theta)`
or `Delta=2absin 2 theta`
`:. Delta_("max")=2ab=8`