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In the standard equation of hyperbola, $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ where $b^2=a^2(e^2-1)$

If I were to draw the graph of hyperbola what would it represent in the graph? As $a$ represents the distance of vertex from the origin.

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    $\begingroup$ $\pm b/a$ are the slopes of the hyperbola's asymptotes. (Note: Whereas $a$ is called the "transverse" (semi)axis of the hyperbola, $b$ is the "conjugate" (semi)axis, which is the transverse (semi)axis of the conjugate hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$$ which has the same center and asymptotes but opens "up and down".) $\endgroup$ – Blue Apr 1 '16 at 6:15
  • $\begingroup$ @Blue I don't think that really answered the question . I don't see how does it represent anything on the graph :/ Well, maybe it did as amd explained. It's just that I think there should be something more then that $\endgroup$ – brainst Apr 1 '16 at 6:19
  • $\begingroup$ Why not trying it ? geogebra Basically it's not a cycle, and looks like an hourglass $\endgroup$ – Aseed Apr 1 '16 at 6:25
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This is the length of the segment perpendicular to the major axis from vertex to either asymptote. Thus, as Blue notes, the asymptotes have slope $\pm\frac b a$.

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  • $\begingroup$ that's it? that property doesn't look like fun :( $\endgroup$ – brainst Apr 1 '16 at 6:22
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    $\begingroup$ @brainst Well, it’s also the length of the semimajor axis of the conjugate hyperbola, but, yeah, that’s about it. $\endgroup$ – amd Apr 1 '16 at 6:40

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