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Find the eccentricity $e$ of the conic $$S \equiv 39x^2+11y^2-96xy+14x+2y-34=0.$$

My try:

Comparing with general second degree conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ we have

$a=39$, $b=11$, $2h=-96$, $2g=14$, $2f=2$ and $c=-34$

The determinant $$\Delta=\begin{vmatrix} 39 & -48&7 \\ -48& 11 & 1\\ 7&1 & -34 \end{vmatrix} \ne 0$$ and

$h^2-ab \gt 0$, and hence the conic is a Hyperbola.

Now since there is $xy$ term one way is to rotate the axes to make $xy$ term zero by angle $$\tan(2\theta)=\frac{-96}{28}.$$ But the algebra is very tedious. Is there any other approach?

Matthew Conroy
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Ekaveera Gouribhatla
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    The shortcut: https://en.wikipedia.org/wiki/Eccentricity_(mathematics)#Values –  Jan 15 '16 at 19:16

2 Answers2

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The formula derived in this answer is $$ e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}} $$ Setting $A=39,B=-96,C=11,D=14,E=2,F=-34$ gives $$ \begin{align} e^2&=\frac{2\cdot100}{-50+100}=4\\ e&=2 \end{align} $$ since $\sigma=\operatorname{sign}\left(F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\right)\right)=\operatorname{sign}\left(-250000\right)=-1$

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robjohn
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Original Conic

a1

First eliminate the offset terms (factors of $x$ and $y$ alone) by finding the "center".

$$\left. \begin{align} \frac{{\rm d}}{{\rm d} x} (39x^2+11y^2-96xy+14x+2y-34)&=0 \\ \frac{{\rm d}}{{\rm d} y} (39x^2+11y^2-96xy+14x+2y-34)&=0 \end{align} \right\} \; \begin{aligned} x & = \frac{1}{15} \\ y & = \frac{1}{5} \end{aligned} $$

Now we substitute $$ \begin{align} x & \rightarrow x + \frac{1}{15} \\ y & \rightarrow y + \frac{1}{5} \end{align} $$

The conic is now $$ 39 x^2 - 96 x y + 11 y^2 - \frac{100}{3} = 0 $$

a2

Now lets find the rotation angle $\theta$ that gets rid of the $xy$ terms

One way to do this is to apply a rotation $$R = \begin{vmatrix} \cos\theta & -\sin \theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix}$$ to the homogeneous coordinates $(x,y,1)$ transforming the conic equation from

$$ 39 x^2 - 96 x y + 11 y^2 - \frac{100}{3} = 0 \rightarrow \\ \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} ^ \top \begin{vmatrix} 39 & -\frac{96}{2} & 0 \\ - \frac{96}{2} & 11 & 0 \\ 0 & 0 & -\frac{100}{3} \end{vmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$

to

$$\begin{pmatrix} x \\ y \\ 1 \end{pmatrix} ^ \top \begin{vmatrix} \cos\theta & -\sin \theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix}^\top \begin{vmatrix} 39 & -\frac{96}{2} & 0 \\ - \frac{96}{2} & 11 & 0 \\ 0 & 0 & -\frac{100}{3} \end{vmatrix} \begin{vmatrix} \cos\theta & -\sin \theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$

$$ \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} ^ \top \begin{vmatrix} 28 \cos^2 \theta-96 \sin\theta\cos\theta+11 & -96 \cos^2\theta-28 \sin\theta\cos\theta+48 & 0 \\ \boxed{-96 \cos^2\theta-28 \sin\theta\cos\theta+48} & 28\sin^2\theta + 96 \sin\theta\cos\theta+11 & 0 \\ 0 & 0 & -\frac{100}{3} \end{vmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$

$$ -96 \cos^2\theta-28 \sin\theta\cos\theta+48 = 0 \\ 96 \left( \frac{1}{2} + \frac{\cos 2\theta}{2} \right) + 28 \frac{\sin 2\theta}{2} = 48 \\ 48 \cos(2\theta) + 14 \sin(2 \theta) = 0 \\ \tan(2\theta) + \frac{24}{7} =0 \\ \theta = -\frac{1}{2} \arctan\left( \frac{24}{7} \right) $$

This angle transforms the conic with rotated cordinates into

$$ Ax^2 + B y^2 + C =0 \\ (75) x^2 + (- 26) y^2 + (- \frac{100}{3}) = 0 $$

a3

Lastly for an ellipse with major radius $r_x$ and minor radius $r_y$ the eccentricity is defined as $$\epsilon = \sqrt{1-\frac{r_y^2}{r_x^2}}$$

Now we convert the conic $Ax^2+By^2+C=0$ into $\left( \frac{x}{r_x}\right)^2 + \left(\frac{y}{r_y}\right)^2 = 1$ form in order to get the major and minor dimensions $r_x$ and $r_y$.

This is done with $$ \left. \frac{x^2}{-C/A} + \frac{y^2}{-C/B} = 1 \right\} \begin{aligned} r_x^2 & =-C/A=\frac{4}{9} \\ r_y^2 & = -C/B =-\frac{50}{39} \end{aligned}$$

The eccentricity is thus

$$ \epsilon = \sqrt{1 - \frac{r_y^2}{r_x^2}} = \sqrt{1-\frac{A}{B}}= \sqrt{1+\frac{75}{26}} = 1.9709427654336\ldots $$

NOTE: That only the coefficients $A$ and $B$ are used in the eccentricity calculation. Different values of $C$ would scale the radii differently but their ratio depends on $A$ and $B$ only.

There is a co-eccentricity defined here as $\epsilon^\star = \sqrt{1-\frac{B}{A}} = 1.160459679035 \ldots$

John Alexiou
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  • Where does $C$ get used to compute $\epsilon$? Just knowing $A$ and $B$, the eccentricity could be $2$, for $C\lt0$, or $\frac2{\sqrt3}$, for $C\gt0$. – robjohn Jan 15 '16 at 22:50
  • it is not used. I just included it here for completeness. $C$ is like a size/scale parameter, and $A$, $B$ dictate the shape. – John Alexiou Jan 16 '16 at 01:00
  • Not so. $75x^2-26y^2-1=0$ has $\epsilon=2$, whereas $75x^2-26y^2+1=0$ has $\epsilon=\frac2{\sqrt3}$. – robjohn Jan 16 '16 at 01:59
  • @robjohn I disagree. Look at my edit, which shows the derivation of eccentricity for a conic more rigorously. Where did you get your eccentricity values from. Note also that the second equation above is equivalent to $-75 x^2 + 26 y^2 -1 =0$ which should different eccentricity not because $C$ is different but because what is considered a major and a minor radius is different (interchanged). – John Alexiou Jan 16 '16 at 16:30
  • The equations in my last comment are links to the curves plotted from those equations. They clearly differ in eccentricity. You cannot tell the eccentricity of a hyperbola just knowing the second order coefficients. The idea of a formula is to be able to use it without plotting and looking to see which axis is major and which is minor. The formula used in my answer gives the eccentricity given only the coefficients (but all of the coefficients). – robjohn Jan 17 '16 at 02:58