Question

Prove that the squares of the semi-axes of the conic $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ are $2△÷\left\{\right(ab-h^2\left)\right(a+b\pm \sqrt{(a-b{)}^2+4h^2}\left)\right\},$ where $△$ is the discriminant.

Answer 1

This is the general equation of second degree, hence the equation referred to parallel axes through the center will be
$a{x}^2+2hxy+b{y}^2+\frac{△}{ab-h^2}=0$
and also
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{a+b}{c}$ and $\frac{1}{{\alpha }^2{\beta }^2}=\frac{c^2}{ab-h^2}$
Hence the value of ${\alpha }^2$ and ${\beta }^2$ are
$\frac{1}{2}\frac{c}{ab-h^2}[a+b\pm \sqrt{{(a-b)}^2+4h^2}]$
or $-\frac{1}{2}\frac{△}{{(ab-h^2)}^2}\frac{4(ab-h^2)}{a+b\pm \sqrt{{(a-b)}^2+4h^2}}$
i.e., $-2△÷\left[(ab-h^2)\{a+b\pm \sqrt{{(a-b)}^2+4h^2}\}\right]$