# How to find the eccentricity of this conic?

How to find the eccentricity of this conic?

4(2y-x-3)² - 9(2x+y-1)²=80

My approach :

I rearranged the terms and by comparing it with general equation of 2nd degree, I found that its a hyperbola. Since this hyperbola is not in standard form x²/a² - y²/b² = 1, I don't know how to find its eccentricity.

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The eccentricity should be preserved by rigid motions, right? –  anon Jun 30 '12 at 7:55
What level are you asking this question from,... is this a pre-calculus course? Just so the responses and terminology would be more appropriate for your level. If so, you might want to tag your question with pre-calculus or the related as well. –  night owl Jun 30 '12 at 8:29

First make the following change of coordinates: $$u=\frac{x-2y}{\sqrt{3}}, \ v=\frac{2x+y}{\sqrt{3}}.$$ With these coordinates the canonical basis $e_1=(1,0), e_2=(0,1)$ is transformed into $e_1'=\frac{(1,2)}{\sqrt{3}}, e_2'=\frac{(-2,1)}{\sqrt{3}}$ which is clearly an orthonormal basis. The equation now reads: $$4(-\sqrt{3}u-3)^2-9(\sqrt{3}v-1)^2=80,$$ i.e. $$\frac{(u+\sqrt{3})^2}{a^2}-\frac{(v-1/\sqrt{3})^2}{b^2}=1.$$ with $$a^2=20/3>b^2=80/27.$$ So, the eccentricity is $$e=\sqrt{1+b^2/a^2}=\sqrt{1+4/9}=\sqrt{13}/3.$$

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Thank you O'Mercy. –  Bazinga Jun 30 '12 at 8:43