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# Need Help ASAP!

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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:

$$y\ '=\pm\ tan(30°)$$

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

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

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

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

$$\frac{b}{a}=tan(30°)$$

$$\frac{b}{a}= \frac{1}{3}\sqrt{3}\\ \frac{a}{b}=\pm \sqrt{3}$$

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

!

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