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How to find the eccentricity of this conic?


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 $x^2/a^2-y^2/b^2= 1$, I don't know how to find its eccentricity.

Please guide me.

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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
up vote 3 down vote accepted

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

HINT: The presence of $x y $ term shows that the axes are not going to be nicely parallel to x- and y- axes. There is a formula like $ \tan 2 \theta = 2B/(A-C) $ to find angle of rotation to align along x- and y-.

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