<s>[INST] Ellipse $3x^2-x+6xy-3y+5y^2=0$: what are the semi-major and semi-minor axes, displacement of centre, and angle of incline? Given the ellipse
$$3x^2-x+6xy-3y+5y^2=0$$
find the following:


*

*semi-major axis, $a$

*semi-minor axis, $b$

*displacement of centre from origin (or coordinates of centre of ellipse $(h,k)$)

*angle of incline $\theta$ (or $\tan\theta$).


This can of course be done systematically by equating coefficients with 
$$\frac{((x-h)\cos\theta-(y-k)\sin\theta)^2}{a^2}+\frac{((x-h)\sin\theta+(y-k)\cos\theta)^2}{b^2}=1$$ 
which is derived by applying the rotation and translation parameters to the standard ellipse equation. Perhaps someone could complete the solution?
Also, can it be solved in any other way, and in particular, without using trigonometric 
ratios?
(NB - this problem came about whilst trying to solve another problem on MSE here
How to Solve this Arithmetic Progression Question? which gave rise to the Diophantine Equation above; assuming instead that $x,y$ can take on non-integer values the equation describes a rotated and translated ellipse)
 [/INST] Here's another method suggested by a friend of mine. 
Keep the solution of the centre of the ellipse identified earlier as $(-\frac13,\frac12)$ .   
Change the centre of the ellipse such to the origin. Equation of translated ellipse is:
$$E':\qquad  3x^2+6xy+5y^2-\frac7{12}=0\qquad \cdots (1)$$
Consider a circle at the same centre as the ellipse (i.e. the origin) with equation
$$C': \qquad x^2+y^2=r^2\qquad \qquad \qquad \qquad \cdots (2) $$
At intersection of $E'$ and $C$:
$$(1)-3\times(2):\\
6xy+2y^2=\frac7{12}-3r^2$$
Substitute in $(1)$:
$$\begin{align}
6\sqrt{r^2-y^2}y&=-\left[2y^2+\left((3r^2-\frac7{12}\right)\right]\\
36(r^2-y^2)y^2&=4y^4+4\left(3r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2\\
40y^4+4\left(-6r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2&=0
\end{align}$$
For tangency,
$$\begin{align}
4^2\left(6r^2+\frac7{12}\right)^2&=4(40)\left(3r^2-\frac7{12}\right)^2\\
6r^2+\frac7{12}&=\sqrt{10}\left(3r^2-\frac7{12}\right)\\
3r^2(2-\sqrt{10})&=\frac7{12}(-1-\sqrt{10})\\
r^2&=\frac7{36}\frac{\sqrt{10}+1}{\sqrt{10}-2}=\frac7{72}(4+\sqrt{10})\\
r&=\sqrt{\frac7{72}(4+\sqrt{10})}\qquad \text{(semi-major axis)}\qquad \blacksquare
\end{align}$$
Correspondingly, 
$$\qquad\qquad r=\sqrt{\frac7{72}(4-\sqrt{10})}\qquad \text{(semi-minor axis)}\qquad \blacksquare$$
The incline of the ellipse can be calculated easily from values of $x,y$ obtained by substituting values of $r$. 
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