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# 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. x^{2}+ 4y^{2} = 4 and rectangle both lie in the first quadrant.

## Answer: The desired ellipse is x^{2}/16 + 3/4 × y^{2} = 1 or x^{2 }+ 12y^{2 }= 16

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 passes 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 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/b^{2 }= 1

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

b^{2 }= 4/3

### Hence the equation of the desired ellipse is x^{2}/16 + 3/4 × y^{2} = 1 or x^{2 }+ 12y^{2 }= 16

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