Question-45

An ellipse can be represented by a quadratic form whose coefficient matrix is positive definite. The general equation is given by: \[ \begin{bmatrix} x & y \end{bmatrix}A \begin{bmatrix} x\\ y \end{bmatrix} = 1 \] Visually:

Consider the ellipse given by:

\[ \begin{equation*} 5x^{2} +8xy+5y^{2} =1 \end{equation*} \]

Select all true statements concerning the major axis of this ellipse.

• A unit vector along the major axis of the ellipse is \(\displaystyle \left(\frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right)\).
• A unit vector along the major axis of the ellipse is \(\displaystyle \left(\frac{-1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right)\).
• The length of the major axis of the ellipse is \(\cfrac{2}{3}\).
• The length of the major axis of the ellipse is \(\displaystyle 2\).

Hint

Diagonalize the coefficient matrix corresponding to the quadratic form.

Answer

• A unit vector along the major axis of the ellipse is \(\displaystyle \left(\frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right)\).
• A unit vector along the major axis of the ellipse is \(\displaystyle \left(\frac{-1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right)\).
• The length of the major axis of the ellipse is \(\cfrac{2}{3}\).
• The length of the major axis of the ellipse is \(\displaystyle 2\).

Solution

Step-1

The coefficient matrix corresponding to the quadratic form is:

\[ \begin{equation*} A=\begin{bmatrix} 5 & 4\\ 4 & 5 \end{bmatrix} \end{equation*} \]

Let us now find the eigenvalues and eigenvectors of \(\displaystyle A\):

\[ \begin{equation*} \begin{aligned} |A-\lambda I| & =\lambda ^{2} -10\lambda +9 \end{aligned} \end{equation*} \]

We get \(\displaystyle \lambda =1,9\). Now for the eigenvectors:

Step-2

For \(\displaystyle \lambda _{1} =1\):

\[ \begin{equation*} \begin{bmatrix} 4 & 4\\ 4 & 4 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \end{equation*} \]

Any non-zero multiple of \(\displaystyle ( -1,1)\) is an eigenvector. A unit eigenvector in this direction is \(\displaystyle \frac{1}{\sqrt{2}}( -1,1)\).

For \(\displaystyle \lambda _{2} =9\):

\[ \begin{equation*} \begin{bmatrix} -4 & 4\\ 4 & -4 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \end{equation*} \]

Any non-zero multiple of \(\displaystyle ( 1,1)\) is an eigenvector. A unit eigenvector in this direction is \(\displaystyle \frac{1}{\sqrt{2}}( 1,1)\).

The major axis lies along the eigenvector with the smallest eigenvalue, here \(\displaystyle \lambda _{2} =1\), namely \(\displaystyle \left(\frac{-1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right)\). The length of the major axis is given by \(\displaystyle \frac{2}{\sqrt{\lambda _{2}}} =2\). Let us visualize this ellipse: