Suppose I have an ellipse/hyperbola rotated about the origin by some angle $\theta$. Am I right in saying that the following general process will find the eccentricity $e$ of these conics?
- Find $a$, the length of the semi-major axis. For ellipse, the major axis is the longest diameter, for the hyperbola, the major axis is the chord between the vertices.
- Find $b$, the length of the semi-minor axis.
- Then, $b^2 = a^2(1-e^2)$ for an ellipse, and $b^2 = a^2(e^2 – 1)$ for a hyperbola, allowing you to solve for the eccentricity $e$.
Then, in general, the focii are the two points that lie a distance $ae$ away from the origin, along the major axis.
Further, in general, the directrices are the lines parallel to the minor axis, at a distance $\frac{a}{e}$ away from the origin.
Best Answer
The equation $$ax^2+2hxy+by^2=c$$ represents a conic rotated about the axis (depending on $a,b,h,c$) with centre $(0,0)$.
You can always rotate the axis to remove the $xy$ term. This makes it easy to find the focus, length of major axis, etc.
To remove $xy$ term without shifting origin, lets rotate the axis by an angle $\theta$.
Let $X,Y$ represent the new coordinate system. Then, $$x=X\cos\theta-Y\sin\theta$$ $$y=X\sin\theta+Y\cos\theta$$ Substituting and letting coefficient of $XY=0$, you get $$\tan2\theta=\frac{2h}{a-b}$$
For example, lets find $e$ of the following equation (which represents an ellipse): $$4x^2+2y^2+2\sqrt{3}xy=1$$
Raotate the axis by $$\tan2\theta=\frac{2h}{a-b}=\sqrt{3}$$ $$\theta=\frac{\pi}{6}$$
Now, substitute $$x=X\frac{\sqrt{3}}{2}-Y\frac{1}{2}$$ $$y=Y\frac{\sqrt{3}}{2}+X\frac{1}{2}$$ You will get $$5X^2+Y^2=5$$ Now you can find eccentricity using the formula you specified. (Rotating the ellipse or hyperbola will not change the eccentricity and the length of major\minor axis.
This is the graph for the two equations:
(Black represents the new ellipse)