What is the equation of the ellipse that inscribes a rectangle, in which another ellipse of equation x^2 +4y^2 = 4 is inscribed ?
The given ellipse i.e. x2+ 4y2 = 4 and rectangle both lie in the first quadrant.
Answer: The desired ellipse is x2/16 + 3/4 × y2 = 1 or x2 + 12y2 = 16
It is a special problem of an ellipse inscribed in another ellipse.
Let us proceed with the solution to the above problem.
Firstly, let us denote the given coordinate (4,0) as A
The ellipse circumscribing the rectangle passes through the point (4, 0),
so its equation is-
x2/16 + y2/b2 = 1, where b is unknown.
To calculate b, we use the diagonal coordinate of the rectangle, i.e. (2,1)
==> thus, (2,1) lies on the ellipse.
Applying (2,1) in the ellipse equation, we get-
==> 4/16 + 1/b2 = 1
1/b2 = 1 - 1/4 = 3/4
b2 = 4/3
Hence the equation of the desired ellipse is x2/16 + 3/4 × y2 = 1 or x2 + 12y2 = 16