Find the eccentricity of the conic given by: (xtan⁡10∘+ytan⁡20∘+tan⁡30∘)(xtan⁡120∘+ytan⁡220∘+tan⁡320∘)+2018=0

Equating to 0 the expressions inside the parentheses we get the equations of two lines, which are the asymptotes of the hyperbola: xtan⁡10∘+ytan⁡20∘+tan⁡30∘=0, −xtan⁡60∘+ytan⁡40∘+tan⁡320∘=0, where I used tan⁡120∘=−tan⁡60∘ and tan⁡220∘=tan⁡40∘ But these lines are perpendicular, because tan⁡10∘tan⁡60∘=tan⁡20∘tan⁡40∘ (remembering that tan⁡60∘=1tan⁡30∘). Hence this is a rectangular hyperbola and its eccentricity is 2.

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Sereinserenormg

Answered question

2022-01-26

Find the eccentricity of the conic given by:

(x {\tan 10}^{\circ} + y {\tan 20}^{\circ} + {\tan 30}^{\circ}) (x {\tan 120}^{\circ} + y {\tan 220}^{\circ} + {\tan 320}^{\circ}) + 2018 = 0

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ataill0k

Beginner2022-01-27Added 18 answers

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}^{\circ} + y {\tan 20}^{\circ} + {\tan 30}^{\circ} = 0, - x {\tan 60}^{\circ} + y {\tan 40}^{\circ} + {\tan 320}^{\circ} = 0

,
where I used

{\tan 120}^{\circ} = - {\tan 60}^{\circ}

and

{\tan 220}^{\circ} = {\tan 40}^{\circ}

But these lines are perpendicular, because

{\tan 10}^{\circ} {\tan 60}^{\circ} = {\tan 20}^{\circ} {\tan 40}^{\circ}

(remembering that

{\tan 60}^{\circ} = \frac{1}{{\tan 30}^{\circ}}

).
Hence this is a rectangular hyperbola and its eccentricity is

\sqrt{2}

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