|
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. Please guide me. |
|||||
|
|
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 $$ e=\sqrt{1+b^2/a^2}=\sqrt{1+4/9}=\sqrt{13}/3. $$ |
|||
|
|
