Matematické Fórum

stuart clark · 06. 03. 2013 06:58

The eccentricity of Conic Section represented by the equation $y = x - \frac{1}{x}$

Brano · 07. 03. 2013 13:21 — Editoval Brano (07. 03. 2013 13:22)

First step is to rotate it into the basic form
$\frac{u^2}{a^2}-\frac{v^2}{b^2}$
so we substitute
$x=u\cos\phi+v\sin\phi$
$y=-u\sin\phi+v\cos\phi$
into the
$x^2-xy=1$
and if we want a term containing $uv$ to vanish, we get
$\sin(2\phi)-\cos(2\phi)=0$
so we can take $\phi=\frac{\pi}{8}$
using this we obtain the equation
$u^2\underbrace{\frac{\sqrt{2}+1}{2}}_\frac{1}{a^2}-v^2\underbrace{\frac{\sqrt{2}-1}{2}}_\frac{1}{b^2}=1$
and thus the eccentricity is
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{4+2\sqrt{2}}$

Or one can just use the formula here.

stuart clark · 13. 03. 2013 15:31

Thanks ↑ Brano:

Matematické Fórum

#1 06. 03. 2013 06:58

conic section

#2 07. 03. 2013 13:21 — Editoval Brano (07. 03. 2013 13:22)

Re: conic section

#3 13. 03. 2013 15:31

Re: conic section

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