An ellipse with equation x2a2+y2b2=1 contains the circles (x−1)2+y2=1 and (x+1)2+y2=1. Then the smallest possible area of the ellipse can be expressed in the form kπ. Find k.
+130466 The center of the ellipse will be the origin
The length of the major axis is 4....so a is 1/2 of this = 2
The length of the minor axis = 2sqrt(2)....so.....b is 1/2 of this = sqrt(2)
So....the equation of the ellipse is
x^2 y^2
___ + _________ = 1
2^2 [sqrt(2)]^2
x^2 y^2
___ + _____ = 1
4 2
The area of the ellipse = pi * a * b = pi * 2 * sqrt(2) = 2sqrt(2) pi
So
k = 2sqrt (2)
Here is the graph : https://www.desmos.com/calculator/ovso9le6pj
