4

Find the eccentricity of $21x^2-6xy+29y^2+6x-58y-151=0$.

I tried factoring but failed. What are the ways to calculate it, short as well as long methods are welcomed.

Are there any better methods to factorise such a quadratic quickly? I am not familiar with eigenvalues and hence cannot carry out any method involving them.

Viktor Vaughn
  • 20,897
SM Sheikh
  • 145
  • 1
    The equation can be written $180,\left({{\left({{y-1}\over{\sqrt{10}}}+{{3,x}\over{\sqrt{10}}} \right)^2}\over{9}}+{{\left({{x}\over{\sqrt{10}}}-{{3,\left(y-1 \right)}\over{\sqrt{10}}}\right)^2}\over{6}}-1\right)=0$ from which you can read off the semi-major and semi-minor axes and get $e=\frac1{\sqrt{3}}.$ – Jan-Magnus Økland Apr 01 '20 at 12:39
  • It is amazing to see this. But it would be more if you could show how you come up with this. – SM Sheikh Apr 01 '20 at 13:56
  • If all you want is a formula, see https://en.wikipedia.org/wiki/Conic_section#Eccentricity_in_terms_of_coefficients. – amd Apr 01 '20 at 21:06

2 Answers2

6

To explain my comment, as requested:

From solving

$\frac{\partial}{\partial x}(21x^2-6xy+29y^2+6x-58y-151)=0,\frac{\partial}{\partial y}(21x^2-6xy+29y^2+6x-58y-151)=0$ the center is $(0,1).$

$x'=x,y'=y-1$

Now consider $21x'^2-6x'y'+29y'^2-180=0$ which is centered at the origin $(x',y')=(0,0).$

The angle this is rotated is given by $\tan(2\theta)=\frac{-6}{21-29}=\frac34$ or $\cos(\theta)=\frac3{\sqrt{10}},\sin(\theta)=\frac1{\sqrt{10}}.$

$x''=\frac3{\sqrt{10}} x'+\frac1{\sqrt{10}} y',$

$y''=-\frac1{\sqrt{10}} x'+\frac3{\sqrt{10}} y'$

with inverse

$x'=\frac3{\sqrt{10}} x''-\frac1{\sqrt{10}} y'',$

$y'=\frac1{\sqrt{10}} x''+\frac3{\sqrt{10}} y''$

Substituting we get $20x''^2+30y''^2-180=0,$ which is $$(\frac{x''}{3})^2+(\frac{y''}{\sqrt{6}})^2 -1=0.$$

The rest is just back-substituting and a scaling.

To answer the question $a=3,b=\sqrt{6}.$ $e=\sqrt{1-\frac{b^2}{a^2}}=\frac1{\sqrt{3}}.$

Jan-Magnus Økland
  • 6,019
3

The most general equation of the second degree has the form: $Ax^2+Bxy+Cy^2+Dx+Ey+F=0\; \; (A,B,C \; \text{not all zero}).$ You can check this link about Rotation of Conic Sections.

The given conic represents an rotated ellipse since $B^2-4AC=(-6)^2-4(21)(29)<0$. (The conic would have been circle if B=0 and A=C).

Through a change of coordinate you can get the above equation in the standard form by substituting $$y=x'\text{sin}\theta+y'\text{cos}\theta$$ and $$x=x'\text{cos}\theta-y'\text{sin}\theta.$$ Once you get the standard form, you only have to apply direct formula for finding eccentricity. I hope you can work from here, for more details the above link will help you, complete procedure has been described there.

Or there's a direct formula here for finding eccentricity. Using the formula, I got : $e^2=0.3334$. You can calculate $e$ (eccentricity) accordingly.

Nitish Kumar
  • 1,588