Transforming the Ellipse

We are given an ellipse with semi-minor axis = (2, 1) and semi-major axis = (-2, 4). Hence, points on this ellipse can be written as

$X = \cos (t) \begin{bmatrix} 2 \\ 1 \end{bmatrix} + \sin (t) \begin{bmatrix} -2 \\ 4 \end{bmatrix}$

where $X = [ x , y ]^T$

In compact form,

$X = \begin{bmatrix} 2 & -2 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} \cos (t) \\ \sin (t) \end{bmatrix}$

Now each position vector is pre-multiplied by the matrix A, hence define Y as

$Y = A X = \begin{bmatrix} 0.5 & 0.75 \\ -0.25 & 1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} \cos (t) \\ \sin (t) \end{bmatrix}$

Performing the matrix multiplication, this results in

$Y = \begin{bmatrix} 1.75 & 2 \\ 0.5 & 4.5 \end{bmatrix} \begin{bmatrix} \cos (t) \\ \sin (t) \end{bmatrix}$

Solving for the vector of the cosine and sine in terms of Y, by matrix inversion, we get

$\begin{bmatrix} \cos (t) \\ \sin (t) \end{bmatrix} = \frac {1} {6.875} \begin{bmatrix} 4.5 & -2 \\ -0.5 & 1.75 \end{bmatrix} Y$

Now since

$\cos (t)^2 + \sin (t)^2 = \begin{bmatrix} \cos (t) & \sin (t) \end{bmatrix} \begin{bmatrix} \cos (t) \\ \sin (t) \end{bmatrix} = 1$

It follows that

$Y^T B Y = 1 ................(*)$

where

$B = \frac {1} {6.875^2} \begin{bmatrix} 4.5 & -0.5 \\ -2 & 1.75 \end{bmatrix} \begin{bmatrix} 4.5 & -2 \\ -0.5 & 1.75 \end{bmatrix}$

Therefore,

$B = \frac {1} {6.875^2} \begin{bmatrix} 20.5 & -9.875 \\ - 9.875 & 7.0625 \end{bmatrix}$

Now it is known that the length of the semi-minor and semi-major axis for the ellipse described by equation (*) are the square roots of reciprocals of the eigenvalues of matrix $B$ . Since this is a 2 x 2 matrix, it is very easy to find the eigenvalues; they are simply the roots of this equation,

$(x - 20.5 )(x - 7.0625) - (9.875)^2 = 0$

divided by the constant $(6.875)^2$ . The above quadratic equation becomes,

$x^2 - 27.5625 x + 47.265625 = 0$

By the quadratic formula,

$x_1 = 1.837329933$

$x_2 = 25.72517$

Therefore, the eigenvalues are

$\lambda_1 = 1.837329933 / (6.875)^2 = 0.038872434$

$\lambda_2 = 25.72517 / (6.875)^2 = 0.544268$

Hence the length of the semi-major axis is

$a = \frac {1} {\sqrt {0.038872434} } = 5.072$