How to find eccentricity of the conic $3x^2+3y^2-2xy-2=0$?

Question

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 $e=\sqrt{1-b^2/a^2}$?

Answer 1

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 $$e=\frac{1}{\sqrt{2}}$$

Answer 2

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