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For a certain hyperbola \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]where $a > b,$ the angle between the asymptotes is $60^\circ.$ Find $\frac{a}{b}.$

 Aug 4, 2019
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For a certain hyperbola \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]where a > b, the angle between the asymptotes is \( 60^\circ\). Find \(\frac{a}{b}\).

 

Hello Guest!

 

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

\(y=b\cdot \sqrt{\frac{x^2}{a^2}-1}\)

 

For \(x→(-\ ∞)\) the following applies:

 

\(y\ '=\pm\ tan(30°)\)

 

\(y\ '=b\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{\frac{x^2-a^2}{a^2}}}\cdot\frac{2x}{a^2}\\ y\ '=b\cdot \frac{1}{2}\cdot \sqrt{\frac{a^2}{x^2-a^2}}\cdot\frac{2x}{a^2}\\ y\ '=\frac{bx}{a^2}\cdot \sqrt{\frac{a^2}{x^2-a^2}}\\ y\ '=\frac{b}{a}\cdot \sqrt{\frac{x^2}{x^2-a^2}}\\\)

\(For\ x→(-\ ∞)\ the\ following\ applies:\)

\(y\ '=\frac{b}{a}\cdot \sqrt{\frac{x^2}{x^2-a^2}}=tan(30°)\)

\(For\ x→(-\ ∞):\\ \sqrt{\frac{x^2}{x^2-a^2}}=1\)

\(\frac{b}{a}=tan(30°)\)

\(\frac{b}{a}= \frac{1}{3}\sqrt{3}\\ \frac{a}{b}=\pm \sqrt{3}\)

\(\color{BrickRed}a>b\\ \color{blue} \frac{a}{b}=\sqrt{3}\)

 

laugh  !

 Aug 5, 2019
edited by asinus  Aug 5, 2019
edited by asinus  Aug 5, 2019
edited by asinus  Aug 5, 2019

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