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I'm looking for a proof of this following theorem :

Given $ax^2+bxy+cy^2=k$, to standardize equation of this conic section with coefficient of $x'$ being $a'$ and coefficient of $y'$ being $b'$, then $a'$ and $b'$ are the solutions to this equation:

$$m^2-(a+c)m+\left( ac-\frac{b^2}{4} \right) = 0$$

M.G
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Erfan Ahmadi
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1 Answers1

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Let $R(\theta)= \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$ such that $ \begin{pmatrix} x' \\ y' \end{pmatrix}=R(\theta) \begin{pmatrix} x \\ y \end{pmatrix}$.

\begin{align*} ax^2+bxy+cy^2 &= \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ \begin{pmatrix} a' & 0 \\ 0 & b' \end{pmatrix} &=R(\theta)^T \begin{pmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{pmatrix} R(\theta) \\ &= \begin{pmatrix} \frac{a+c}{2}+\frac{a-c}{2}\cos 2\theta-\frac{b}{2} \sin 2\theta & \frac{b}{2}\cos 2\theta+\frac{a-c}{2}\sin 2\theta \\ \frac{b}{2}\cos 2\theta+\frac{a-c}{2}\sin 2\theta & \frac{a+c}{2}+\frac{c-a}{2}\cos 2\theta+\frac{b}{2} \sin 2\theta \end{pmatrix} \\ \end{align*}

Now \begin{align*} \tan 2\theta &= \frac{b}{c-a} \\ a'+b' &= a+c \\ a'b' &= \left( \frac{a+c}{2} \right)^2- \left( \frac{c-a}{2}\cos 2\theta+\frac{b}{2}\sin 2\theta \right)^2 \\ &= \left( \frac{a+c}{2} \right)^2- \left( \frac{b}{2\tan 2\theta}\cos 2\theta+\frac{b}{2}\sin 2\theta \right)^2 \\ &= \left( \frac{a+c}{2} \right)^2- \frac{b^2}{4\sin^2 2\theta} \\ &= \left( \frac{a+c}{2} \right)^2- \frac{b^2(1+\tan^2 2\theta)}{4\tan^2 2\theta} \\ &= \left( \frac{a+c}{2} \right)^2- \frac{(c-a)^2+b^2}{4} \\ &= ac-\frac{b^2}{4} \end{align*}

Note briefly

$a'$ and $b'$ are the eigenvalues of $ \begin{pmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{pmatrix}$, that is $$\det \begin{pmatrix} a-m & \frac{b}{2} \\ \frac{b}{2} & c-m \end{pmatrix} =0$$

Ng Chung Tak
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