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QUESTION {For each of the following quadratic forms, use
eigenvalues and eigenvectors to rotate the axes in order to
identify what type of conic it represents.}

\medskip

\hspace{1cm}(a) $9x^2 - 4xy + 6y^2 - 10x - 20y - 5 = 0$,

\hspace{1cm}(b) $3x^2 - 8xy - 12y^2 - 30x - 64y = 0$,

\hspace{1cm}(c) $4x^2 -20xy + 25y^2 - 15x - 6y = 0$,

\hspace{1cm}(d) $9x^2 +12xy + 4y^2 -52 = 0$.



ANSWER In matrix form the equation is

$\textbf{x}^tA\textbf{x}-\left[\begin{array}{cc}10&amp;20\end{array}\right]\textbf{x}-5=0$

where

$A=\left[\begin{array}{cc}9&amp;-2\\-2&amp;6\end{array}\right]
\hspace{1.5cm}
\textbf{x}=\left[\begin{array}{c}x\\y\end{array}\right]$

The eigenvalues of $A$ are 5 (eigenvector col(1,2)) and 10
(eigenvector col(2,-1)). There are several different orthogonal
matrices which can be used, half have det=1 so are pure rotation
matrices ( the others have det=-1 so reverse orientation as well).
Any one will do to identify the type of conic. Putting
$\textbf{x}=P\underline{\xi}$ where

$P=\left[\begin{array}{cc}\frac{2}{\sqrt{5}}\frac{1}{\sqrt{5}}\\-\frac{1}{\sqrt{5}}\frac{2}{\sqrt{5}}\end{array}\right]\
\hspace{1.5cm}
\underline{\xi}=\left[\begin{array}{c}\xi\\\eta\end{array}\right]$

the equation becomes

$10\xi^2+5\eta^2-10\sqrt{5}\eta-5=0$,\ \ or
$2\xi^2+\eta^2-2\sqrt{5}\eta-1=0$

It is also possible to get any of

\begin{tabular}{ll}
$2\xi^2+\eta^2+2\sqrt{5}\eta-1=0$&amp;$2\eta^2+\xi^2-2\sqrt{5}\xi-1=0$\\
$2\eta^2+\xi^2+2\sqrt{5}\xi-1=0$
\end{tabular}

This may also be written
$\frac{\xi^2}{3}+\frac{(\eta-\sqrt{5})^2}{6}=1$ which is an
ellipse.

The equation $3x^2-8xy-12y^2-30x-64y=0$ has eigenvalues 4
(eigenvector col(4,-1)) and -13 (eigenvector col(1,4)). The
equation can be transformed to :

$4\xi^2-13\eta^2-\frac{56}{\sqrt{17}}\xi-\frac{286}{\sqrt{17}}\eta=0$

or to
$4(\xi-\frac{7}{\sqrt{17}})^2-13(\eta+\frac{11}{\sqrt{17}})^2=-81$

which is a hyperbola.

The equation $4x^2-20xy+25y^2-15x-6y=0$ has eigenvalues )
(eigenvector col(5,2)) and 29 (eigenvector col(-2,5)).The equation
can be transformed to :

$29\xi^2-\frac{87}{\sqrt{29}}\eta=0$ and to
$\eta^2-\frac{3}{\sqrt{29}}\eta=0$

which is a parabola.

The equation $9x^2-20xy+4y^2-52=0$ has eigenvalues 0 (eigenvector
col(2,-3)) and 29 (eigenvector col(3,2)). The equation can be
transformed to :

$\eta^2=4$

which is a pair of parallel straight lines.


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