Parallelogram circumscribed ellipse of minimum area

Geometry Level 4

The four vertices of a parallelogram are given by: $A(0, 3), B(10, 8), C(12, 12), D(2, 7)$ . We want to draw an ellipse that passes through these four points thus circumscribing the parallelogram. There is an infinite number of such ellipses. Find the ellipse of minimum area , and compute that minimum area. If the minimum area can be expressed as $n \pi$ , for an integer $n$ , then enter $n$ as your answer.

12 15 18 24

1 solution

Hosam Hajjir
Apr 20, 2021

To find the minimum area ellipse passing through the vertices of the parallelogram, we're going to transform the parallelogram into a unit square whose minimum area ellipse is known (it is the circle passing through its vertices). Let's transform the parallelogram into a square of side length $2$ centered at the origin, that is, its vertices are $(1, 1), (-1, 1), (-1, -1), (1, -1)$ .

The center of the parallelogram gets shifted to the origin. The center of the parallelogram is at the midpoint of $AC$ , i.e. at $\mathbf{P} = (6, 7.5)$ .

Hence, we can write the following matrix equation:

$\mathbf{y} = T (\mathbf{x} - \mathbf{P} )$

Using $\mathbf{x}_1 = (0, 3)$ with its image $\mathbf{y}_1 = (1, 1)$ , and $\mathbf{x}_2 = (10, 8)$ with its image $\mathbf{y}_2 = (-1, 1)$ , and subsituting this into the above equation, we get

$\begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} = T \begin{bmatrix} -6 && 4 \\ -4.5 && 0.5 \end{bmatrix}$

This can be solved for $T$ by inverting the matrix on the right hand side and right multiplying the matrix on the left hand side by this inverse.

The inverse of the matrix on the right hand side is

$\begin{bmatrix} -6 && 4 \\ -4.5 && 0.5 \end{bmatrix}^{-1} = \dfrac{1}{15} \begin{bmatrix} 0.5 && -4 \\ 4.5 && -6 \end{bmatrix}$

Therefore,

$T = \dfrac{1}{15} \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} \begin{bmatrix} 0.5 && -4 \\ 4.5 && -6 \end{bmatrix} = \dfrac{1}{15} \begin{bmatrix} -4 && 2 \\ 5 && -10 \end{bmatrix}$

The equation of the circle passing through the vertices of the square is

$\mathbf{y}^T \mathbf{y} = R^2 = ( \sqrt{2} )^2 = 2$

Substituing $\mathbf{y} = T ( \mathbf{x} - P )$ , yields

$( \mathbf{x} - P)^T T^T T (\mathbf{x} - P) = 2$

i.e.

$(\mathbf{x} - P)^T G (\mathbf{x} - P ) = 1$

where $G = \frac{1}{2} T^T T = \dfrac{1}{450} \begin{bmatrix} 41 && -58 \\ -58 && 104 \end{bmatrix}$

$G$ is positive definite because $41 \gt 0$ and $(41)(104) - (-58)^2 = 900 \gt 0$ , hence the equation above represents an ellipse, as expected. Moreover, the determinant of $G$ is the reciprocal of the square of the product $a b$ of its semi-axes. Now the determinant is

$| G | = \dfrac{ (41)(104) - (-50)^2 }{450^2} = \dfrac{ (30)^2 }{ (450)^2 }$

Therefore, $a b = \dfrac{450}{30} = 15$ , and the area of the minimum-area ellipse is $\pi a b = 15 \pi$ .

Parallelogram circumscribed ellipse of minimum area

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