Question

Prove that the eccentricity of the conic given by the general equation satisfies the relation $\frac{e^4}{1-e^2}+4=\frac{(a+b-2h\cos \omega {)}^2}{(ab-h^2){\sin }^2\omega },$ where $\omega$ is the angle between the axes.

Answer 1

For the general equation of a conic, we have
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{a+b-2h\cos \omega }{{\sin }^2\omega }$
and $\frac{1}{{\alpha }^2{\beta }^2}=\frac{ab-h^2}{{\sin }^2\omega }$
hence, ${\alpha }^2+{\beta }^2=\frac{a+b-2h\cos \omega }{ab-h^2}$ ....(1)
and ${\alpha }^2{\beta }^2=\frac{{\sin }^2\omega }{ab-h^2}$ ....(2)
Again as $\frac{e^2}{2-e^2}=\frac{{\alpha }^2-{\beta }^2}{{\alpha }^2+{\beta }^2}$, hence $\frac{e^4}{{(2-e^2)}^2}=\frac{{({\alpha }^2+{\beta }^2)}^2-4{\alpha }^2{\beta }^2}{{({\alpha }^2+{\beta }^2)}^2}$
$\frac{e^4}{{(2-e^2)}^2}=1-\frac{4{\alpha }^2{\beta }^2}{{({\alpha }^2+{\beta }^2)}^2}$
$\frac{4{\alpha }^2{\beta }^2}{{({\alpha }^2+{\beta }^2)}^2}=1-\frac{e^4}{{(2-e^2)}^2}=\frac{4-4e^2}{{(2-e^2)}^2}$
$\frac{{({\alpha }^2+{\beta }^2)}^2}{{\alpha }^2{\beta }^2}=\frac{4-4e^2+e^4}{1-e^2}$
Substituting the values from (1) and (2) and simplifying, we get
$\frac{{(\alpha +\beta -2h\cos \omega )}^2}{(ab-h^2){\sin }^2\omega }=4+\frac{e^4}{1-e^2}$