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Find the equations of major and minor axis of an ellipse $$21x^2-6xy+29y^2+6x-58y-151=0$$ and also eccentricity of an ellipse.

What I tried. Let $S = 21x^2-6xy+29y^2+6x-58y-151$

For coordinate of center $\displaystyle \dfrac{dS}{dx}=0$ and $\displaystyle \dfrac{dS}{dy}=0$. Therefore $42x-6y+6=0$ and $58y-6x-58=0$

Solving it i have the center $(0,1)$.

How do i find the axes?

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    $\begingroup$ What have you tried? Did you try to translate the axes $X=x-0$, $Y=y-1$? $\endgroup$ – Chrystomath Mar 8 '19 at 11:00
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Yes, the center is $(x,y)=(0,1)$, so after letting $X=x$ and $Y=y-1$, the equation of the ellipse can be written as $$21X^2-6XY+29Y^2-180=0.$$ Now, the center is $(X,Y)=(0,0)$ and in order to find the axes we need the eigenvectors of the matrix $$\begin{pmatrix} 21 & -3 \\ -3 & 29 \end{pmatrix}$$ whose characteristic equation is $$(21-z)(29-z)-(-3)^2=(z-20)(z-30)=0.$$ For the eigenvalue $\lambda=20$, the eigenspace is generated by the vector $(3,1)^T$ and therefore one of the axes is $y-1=Y/1=X/3=x/3$, i.e. $y=\frac{x}{3}+1$.

Are you able to find the other one?

As regards the eccentricity, find semi-axis's lengths $a$, $b$ and recall the definition.

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Let $f(x,y)=x^2+(y-1)^2$ $$\max_{S(x,y)=0}{f}=9,\quad \min_{S(x,y)=0}{f}=6$$ Then semi-major and semi-minor axes are $$a=3,\quad b=\sqrt{6}.$$

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  • $\begingroup$ I did not understand How can i find $a$ and $b$ explain me please $\endgroup$ – jacky Mar 8 '19 at 11:28
  • $\begingroup$ @jacky With Lagrange multiplier method. $\endgroup$ – Aleksas Domarkas Mar 8 '19 at 12:10
  • $\begingroup$ can you show here please thanks $\endgroup$ – jacky Mar 8 '19 at 13:45

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