The area enclosed by the ellipse `x^2/a^2+y^2/b^2=1` is :

Verified by Experts

The correct Answer is:

`piab`

Similar Questions

Explore conceptually related problems

Find the area bounded by the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 and the ordinates x=0 and x=ae, where b^2=a^2(1-e^2)" and "e lt 1 .

The product of the perpendiculars from the foci on any tengent to the ellipse x^2/a^2 + y^2/b^2 = 1 is

Find the area of the ellipse x^(2)/a^(2)+y^(2)/b^(2)=1

The area enclosed between the curves y=x^2 and x=y^2 is :

Area of the ellipse x^(2)/9+y^(2)/4=1 is

Find the area of the ellipse (x^(2))/(a^(2))+ (y^(2))/(b^(2))=1 by the method of integration.

Find the area of the smaller region bounded by the ellipse x^(2)/a^(2) + y^(2)/b^(2) =1 and the line x/a + y/b =1 .

Find the maximum area of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 which touches the line y=3x+2.

If a tangent of slope 2 of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 is normal to the circle x^2+y^2+4x+1=0 , then the maximum value of a b is

Area of the greatest rectangle that can be inscribed in the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is :