TITLE: Eccentricity of conic given by a complicated equation with trigonometric coefficients such as $\tan 10^\circ$
QUESTION [5 upvotes]: Find the eccentricity of the conic given by:
$$\left(x\tan 10^\circ+y\tan 20^\circ+\tan 30^\circ\right)\left(x\tan 120^\circ+y\tan 220^\circ+\tan 320^\circ\right)+2018=0$$

What I have tried
$$\bigg(x\tan10^\circ+y\tan 20^\circ+\frac{1}{\sqrt{3}}\bigg)\bigg(\sqrt{3}\; x +y\tan 220^\circ+\tan 320^\circ\bigg)+2018=0$$
$$\begin{align}\Longrightarrow\quad 
&\sqrt{3}x^2+\sqrt{3}xy\tan 20^\circ+x+xy\tan 10^\circ\tan 220^\circ+y^2\tan 20^\circ\tan 220^\circ \\[4pt]
&+\frac{y}{\sqrt{3}}\tan 220^\circ+x\tan 10^\circ\tan 220^\circ+y\tan 20^\circ\tan 320^\circ+\frac{1}{\sqrt{3}}\tan 320^\circ \\[4pt]
&+2018=0
\end{align}$$
How do I solve it? Help me, please.

REPLY [4 votes]: Equating to $0$ the expressions inside the parentheses we get the equations of two lines, which are the asymptotes of the hyperbola:
$$
x\tan 10°+y\tan 20°+\tan 30°=0,\quad
-x\tan 60°+y\tan 40°+\tan 320°=0,
$$
where I used $\tan120°=-\tan60°$ and $\tan220°=\tan40°$.
But these lines are perpendicular, because
$$\tan10°\tan60°=\tan20°\tan40°$$
(you can check that with a calculator, or read the answers to this question, remembering that $\tan60°=1/\tan30°$).
Hence this is a rectangular hyperbola and its eccentricity is $\sqrt2$.