## Greatest and Least Distance of an Ellipse in Quadratic Form From the Origin

Suppose we have a quaratic form
$Ax_1^2+Bx_1x_2+Cx^2_2=k$
.
We can write this in matrix form as
$(x_1,x_2) \left( \begin{array}{ccc} A & B/2 \\ B/2 & C \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}=k$
.
If the eigenvalues are both positive then the curve is an ellipse. The points furthest from the origin will be at either end of the longest axis. The axes are in the directions of the eigenvalues, and the longest axis will be in the direction of the smallest eigenvalue.
We can find the eigenvalues
$\lambda_1 , \: \lambda_2$
and corresponding eigenvectors of the matrix above and express the conic in the canonical form
$\lambda_1 y_1^2+ \lambda_2 y_2^2=C'$
.
The least distance of the curve from the origin will be
$\sqrt{C'/ \lambda'}$
where
$\lambda'$
is the smallest eigenvalue, and the least distance will be
$\sqrt{C'/ \lambda''}$
where
$\lambda''$
is the largest eigenvalue.
Example: Write the conic
$5x_1^2+4x_1x_2+8x_2^2=9$
in canonical form.
We find the eigenvalues and eigenvectors of the matrix
$A=\left( \begin{array}{cc} 5 & 4/2 \\ 4/2 & 8 \end{array} \right)=\left( \begin{array}{cc} 5 & 2 \\ 2 & 8 \end{array} \right)$

\begin{aligned} \left| \begin{array}{cc} 5- \lambda & 2 \\ 2 & 8- \lambda \end{array} \right| & =(5-- \lambda)(8- \lambda )-2^2 \\ &=\lambda^2-13 \lambda-36 \\ & =(\lambda-4)(\lambda-9)=0 \end{aligned}

Hence
$\lambda=4, \: 9$

$\lambda=4$
:
\begin{aligned}\left( \begin{array}{cc} 5-4 & 2 \\ 2 & 8-4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}&=\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}\\ &=\begin{pmatrix}x_1+2x_2\\2x_1+4x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix} \end{aligned}

We can take
$x_1=-2, \: x_2=1$

Normalising gives
$x_1=-\frac{2}{\sqrt{5}}, \: x_2=\frac{1}{\sqrt{5}}$

The first eigenvector is
$\begin{pmatrix}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{pmatrix}$

$\lambda=9$
:
\begin{aligned} \left( \begin{array}{cc} 5-9 & 2 \\ 2 & 8-9 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \left( \begin{array}{cc} -4 & 2 \\ 2 & -1 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} \\ &= \begin{pmatrix}-4x_1+2x_2\\2x_1-x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix}\end{aligned}

We can take
$x_1=1, \: x_2=2$

Normalising gives
$x_1=-\frac{1}{\sqrt{5}}, \: x_2=\frac{2}{\sqrt{5}}$

The second eigenvector is
$\begin{pmatrix}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{pmatrix}$

The matrix of eigenvectors is
$\left( \begin{array}{cc} -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array} \right)$

Now define the transformation
$\mathbf{x}=P \mathbf{y}$
then
$\mathbf{x}^TA \mathbf{x}=9$
becomes
$(P \mathbf{y})^TA(P \mathbf{x})=\mathbf{y}^TP^TAP=\mathbf{y}D\mathbf{y}=9$

We have
$(y_1.y_2)\left( \begin{array}{cc} 4 & 0 \\ 0 & 9 \end{array} \right)\begin{pmatrix}y_1\\y_2\end{pmatrix}=4y_1^2+9y_2^2=9$

The greatest distance is
$\sqrt{9/4}=3/2$
and the least distance is
$\sqrt{9/9}=1$