Finding Eccentricity from the rotating ellipse formula

Question

I see that from a normal ellipse formula, we can acquire the eccentricity via this formula here.

However, for this formula (1):

$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$

When parameter $B = 0$, we would have normal ellipse, and the formula from the link above can be used.

But when $B ≠ 0$, we will have a tilting ellipse, and its eccentricity will change as well. In fact, if we are to find the current eccentricity of the given formula (1), what would be the formula for the eccentricity in this case?

Answer 1

As the translation does not change the eccentricity of a conic, let's us define $x=X+h,y=Y+k$ to get $AX^2+BXY+CY^2=1$

Using Rotation of axes as eccentricity is one of the invariants in rotation,

$$x'^2(A\cos^2t+B\sin t\cos t+C\sin^2t)+x'y'(B\cos2t-(A-C)\sin2t)+y'^2(A\sin^2t-B\sin t\cos t+C\cos^2t)=1$$

To remove the $x'y'$ term, $\tan 2t=\frac{B}{A-C}\implies \frac{\sin2t}{B}=\frac{\cos2t}{A-C}=\pm\frac1{\sqrt{B^2+(A-C)^2}}$

If we take the $'+'$ sign, $\sin2t=\frac B{\sqrt{B^2+(A-C)^2}},\cos2t=\frac {A-C}{\sqrt{B^2+(A-C)^2}}$

Comparing with the standard form we get

$a^2=\frac1{A\cos^2t+B\sin t\cos t+C\sin^2t}$ $=\frac2{A(1+\cos2t)+B\sin2t+C(1-\cos2t)}$ $=\frac{2\sqrt{B^2+(A-C)^2}}{(A+C)\sqrt{B^2+(A-C)^2}+(A-C)^2+B^2}$ $=\frac2{A+C+\sqrt{B^2+(A-C)^2}}$

Similarly, $b^2=\frac2{A+C-\sqrt{B^2+(A-C)^2}}>a^2$

So, $b$ is the semi-major axis, $a$ is the semi-minor axis.

Consequently, the equation of the conic becomes $$\frac{x'^2}{\frac2{A+C+\sqrt{B^2+(A-C)^2}}}+\frac{y'^2}{\frac2{A+C-\sqrt{B^2+(A-C)^2}}}=1$$

$e^2=1-\frac{a^2}{b^2}=1-\frac{A+C-\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}=\frac{2\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}$

If $A+C-\sqrt{B^2+(A-C)^2}=0$ i.e., $4AC=B^2,e^2=1,$ the conic becomes a Parabola.

If $A+C-\sqrt{B^2+(A-C)^2}>0$ i.e., $4AC>B^2,e^2<1,$ the conic becomes an ellipse.

If $A+C-\sqrt{B^2+(A-C)^2}<0$ i.e., $4AC<B^2,e^2>1,$ the conic becomes a hyperbola.

Answer 2

You find the eigenvalues of the matrix $M$ in Robert's answer here, $$M = \begin{bmatrix}A & B/2 \\ B/2 & C\end{bmatrix}.$$

The two eigenvalues of $M$ have an explicit formula since $M$ is a $2\times2$ matrix: $$\lambda = \frac12\left(A+C\pm\sqrt{(A-C)^2+B^2}\right).$$

Knowing these are equal to $1/a^2$ and $1/b^2$, you can find $a$ and $b$ to plug into the eccentricity formula.

Answer 3

If $S(p,q)$ be a focus, $L: ax+by+c=0$ be the directrix and $e$ be the eccentricity of a conic with $P(x,y)$ is any point on the conic,

$$e=\frac{\sqrt{(x-p)^2+(y-q)^2}}{\frac{|ax+by+c|}{\sqrt{a^2+b^2}}}$$

On simplification, $$\{a^2(1-e^2)+b^2\}x^2-2abe^2xy+\{a^2+(1-e^2)b^2\}y^2+()x+()y+()=0$$

Comparing with the given equation, $a^2(1-e^2)+b^2=A, a^2+(1-e^2)b^2=C$ and $-2abe^2=B$

So, we have three unknowns $a,b,e$ with three equations which can be solved easily.

On elimination of $a,b,$ we get $$(B^2-4AC)e^4-4e^2\{B^2+(A-C)^2\}+4\{B^2+(A-C)^2\}=0$$

The value of discriminant is $4^2\{B^2+(A-C)^2\}^2-4(B^2-4AC)4\{B^2+(A-C)^2\}$ $=16\{B^2+(A-C)^2\}\{B^2+(A-C)^2-(B^2-4AC)\}=16(A+C)^2\{B^2+(A-C)^2\}$

So, $e^2=\frac{4\{B^2+(A-C)^2\}\pm4(A+C)\sqrt{B^2+(A-C)^2}}{2(B^2-4AC)}=\frac{2\sqrt{B^2+(A-C)^2}\{\sqrt{B^2+(A-C)^2}\pm(A+C)\}}{B^2-4AC}$

For the '-' sign, this is the same value we reached at in the 1st answer as $B^2-4AC=B^2+(A-C)^2-(A+C)^2=\{\sqrt{B^2+(A-C)^2}+A+C\}\{\sqrt{B^2+(A-C)^2}-A-C\}$

Answer 4

In my answer to a related question, it is given that $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ is simply a rotated and translated version of $$ \small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0 $$ From this, we get the eccentricity of the conic is given by $$ e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}} $$ where $\sigma$ is the sign of $F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\right)$.