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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.

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jacky
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1 Answers1

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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 $\sqrt2$.

Intelligenti pauca
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