# 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. x2/4 + y2 = 1 and rectangle both lie in the first quadrant.

## Answer: The desired ellipse is x^{2}/16+3/4 * y^{2} = 1

It is a special problem of an ellipse inscribed in another ellipse.

**Explanation:**

Let us proceed with the solution to the above problem.

Firstly, let us denote the given coordinate (4,0) as A

Given,

The ellipse circumscribing the rectangle passess through the point (4, 0),

so its equation is-

x^{2}/16+y^{2}/b^{2}=1, where b is unknown.

To calculate b, we use the the diagnol 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/b^{2}=1

1/b^{2}=1-1/4=3/4

b^{2}=4/3