How to find eccentricity of the conic $3x^2+3y^2-2xy-2=0$? How to find eccentricity of the conic $3x^2+3y^2-2xy-2=0$?
How to deal with the $xy$ term?
Can't understand. Help!
Why isn't the standard formula working here for eccentricity of an ellipse which states $e=\sqrt{1-b^2/a^2}$?
 A: Exchanging $x$ and $y$ would give the same equation. This would indicate that the line $x-y=0$ is a line of symmetry, and therefore one of the axes of the ellipse. Therefore so is the line $x+y=0$.
Therefore, we consider a change of variable: let $u=x+y$ and $v=x-y$ so that $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$.
Substituting these into the given equation and simplifying, we have $$\frac{u^2}{2}+v^2=1$$
Now the ellipse is in standard form, with $a^2=2$ and $b^2=1$. Using $b^2 =a^2(1-e^2)$ gives $$e=\frac{1}{\sqrt{2}}$$
A: HINT For conics of the form $$ax^2+2hxy+by^2+cx+dy+e=0$$
$ab-h^2$
the way to determine the nature of the conic is to solve $ab-h^2$
In each case it is given by:


*

*hyperbola : $ab-h^2<0$

*ellipse : $ab-h^2>0$

*parabola : $ab-h^2=0$
Where c represents the length of the semi major axis and d represents the length of the semi minor axis.
To get this form you must first find the centre points by calling the initial equation f(x,y) and making $f_x=0$ and $f_y=0$ and solving from here
for x and y to get the centre point (h,k).
Now you have these the new coordinate values are $u=x-h$ and $v=y-k$.
Then, subbing x and y in terms of u and y back into the initial equation you should find that depending on the nature of the conic, then it should come in the form of:


*

*hyperbola : $\frac{u^2}{c^2}-\frac{v^2}{d^2}=1$

*ellipse : $\frac{u^2}{c^2}+\frac{v^2}{d^2}=1$

*parabola : $u^2=4cv$
and now:


*

*hyperbola: $e=\sqrt{1+\frac{c^2}{d^2}}$

*ellipse: $e=\sqrt{1-\frac{c^2}{d^2}}$

*parabola: $e=1$
NOTE Dont forget the 2h in the original equation
