# Eccentricity of conic given by a complicated equation with trigonometric coefficients such as $\tan 10^\circ$

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.

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 $$\sqrt2$$.