If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.
 A: Let
$$f(x,y)= 21x^2 -6xy +29y^2 +6x-58y-151 $$
$$f_x’(x,y)= 42x-6y+6$$
$$f_y’(x,y)= 6x+58y-58$$
The center $(0,1)$ is obtained via $f’_x=f’_y=0$
and by matching the normal vectors at  vertexes, i.e.
$$f’_x:f’_y=(x-0):(y-1)$$
the equations for the major and minor axes are obtained
$$3y-x-3=0,\>\>\>\>\>y+3x-1=0$$
Then, substitute  them into $f(x,y)= 0$ to get the major vertexes
$(\pm\frac9{\sqrt{10}},1\pm\frac3{\sqrt{10}})$ and the minor vertexes
$(\mp\sqrt{\frac35},1\pm3\sqrt{\frac35})$, and in turn their respective lengths $2a=6$ and $2b=2\sqrt6$. Thus, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}= \frac1{\sqrt3}$.
A: The fact that there is an $xy$ term means that the axes of the conic are rotated from the coordinate axes.  So the first thing I would do is rotate to new $x'y'$ at some angle $\theta$:
$$x= x'\cos(\theta)- y'\sin(\theta), \quad y= x'\sin(\theta)+ y'\cos(\theta)$$
\begin{align}
21x^2&= 21(x'^2\cos^2(\theta)- 2x'y'\sin(\theta)\cos(\theta)+ y'^2\sin^2(\theta))\\
-6xy&= -6(x'^2\cos(\theta)\sin(\theta)+ x'y'(\cos^2(\theta)- \sin^2(\theta))- y'^2\sin(\theta)\cos(\theta))\\
29y^2&= 29(x'^2\sin^2(\theta)+ 2x'y'\sin(\theta)\cos(\theta)+y'^2\cos^2(\theta)\\\\
21x^2- 6xy+ 29y^2&= (21\cos^2(\theta)- 6\cos(\theta)\sin(\theta)+ 29\sin^2(\theta))x'^2\\&\qquad+ (-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta))x'y'\\
&\qquad+ (21\sin^2(\theta)+ 6\sin(\theta)\cos(\theta)+ \cos^2(\theta)y'^2
\end{align}
The point is to eliminate the coefficient of $x'y'$ so we must have $$-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta)= 0$$ or
$$6\sin^2(\theta)+ 69\sin(\theta)\cos(\theta)- 6\cos^2(\theta)= 0$$
We can think of that as the quadratic equation $6X^2+ 69XY- 6Y^2= 0$.  By the quadratic formula $$X= \frac{-69Y\pm\sqrt{4761Y^2+ 144Y^2}}{12}= \frac{-69Y\pm Y\sqrt{4905}}{12}= \frac{1.03}{12}Y.$$
So $\sin(\theta)= \dfrac{1.03}{12}cos(\theta)$ and $\tan(\theta)= \dfrac{1.03}{12}$.
