Finding the eccentricity of a general two degree curve .

Question

Find the eccentricity of 21x²-6xy+29y²+6x-58y-151=0. I tried factoring but failed. What are the ways to calculate it,short as well as long method are accepted.or is there any better method to factorise quickly? .I don't know Eigen value method.

Answer 1

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.

Answer 2

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