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The parametric point on hyperbola is $(a\sec\theta, b\tan\theta)$.

Why is that when the parametric angle $\theta$ is in 1st quadrant, it represents the part of hyperbola in 1st quadrant, and similarly for the 4th quadrant, but in the 2nd and 3rd quadrants the angle $\theta$ and the part of hyperbola it represents get interchanged?

MPW
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DrivenMad
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    For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Nov 16 '23 at 18:17
  • Thanks, I'm new here. – DrivenMad Nov 16 '23 at 18:35
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    Welcome to Math.SE! ... The reason that the parametric angle and the corresponding part of the hyperbola get interchanged is simply that $\tan\theta$ (the point's $y$-coordinate) is negative for $\theta$ in the second quadrant, and positive for $\theta$ in the third quadrant. ... Generally, there's no requirement that the geometry of a parameter match the geometry of the curve being drawn. For instance, we can parameterize a complete unit circle via $(\cos10\theta, \sin10\theta)$, while parameter $\theta$ itself only varies from $0^\circ$ to $36^\circ$. – Blue Nov 16 '23 at 19:08
  • @Blue yep, got it. Thanks. – DrivenMad Nov 16 '23 at 19:17
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    @StutiGupta: As a bit of a follow-up to my comments to your other question ... Note that, with the hyperbola parameterization $(a\sec\theta,b\tan\theta)$, as $\theta$ approaches, reaches, and passes $90^\circ$, the parameterized point approaches, "reaches", and "passes through, coming back from the other side of" the point-at-infinity of one of the asymptotes; likewise for $270^\circ$. So, the weird quadrant-swapping behavior you've asked about is precisely what causes the parameterization to trace the hyperbola as a "continuous loop". – Blue Nov 19 '23 at 01:14
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    @Blue thank you for your help – DrivenMad Nov 19 '23 at 16:59

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