Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$ e=\sqrt{1+b^2/a^2}=\sqrt{1+4/9}=\sqrt{13}/3. $$

share|cite|improve this answer
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-.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.