Slant/obliques Asymptotes of Polar Curves I'm having some trouble understanding the behaviors of polar curves that have slant asymptotes, in particular the polar representations of hyperbolas.
I think my confusion is best illustrated with a specific example. Take the polar curve $r=\frac{6}{1+2\sin \theta}$, which represents a hyperbola with eccentricity $2$ and center at $(0,4)$. I understand that, as a hyperbola, its graph should possess asymptotes that pass through the center point. However, my confusion arises as I attempt to sketch a graph of the polar curve using my basic understanding of the polar coordinate system. From $0≤\theta≤\pi$ I have no conceptual issues; the curve begins at the polar coordinate $(6,0)$, intercepts the $y$-axis at $(2,\frac{\pi}{2})$, and rejoins the $x$-axis at $(6,\pi)$.
From there, it's clear that as $\theta \rightarrow (\frac{7\pi}{6})^-$, $r\rightarrow\infty$; similarly, as $\theta \rightarrow (\frac{7\pi}{6})^+$, $r\rightarrow-\infty$. Given that vertical asymptotes in Cartesian coordinates occur when $$\lim_{x\to a^±}f(x)\rightarrow±\infty$$ and possess equations of the form $$x=a$$ I would intuit that because $$\lim_{\theta\to (\frac{7\pi}{6})^-}r(\theta)\rightarrow\infty$$ and $$\lim_{\theta\to (\frac{7\pi}{6})^+}r(\theta)\rightarrow-\infty$$ the polar curve possesses an oblique asymptote with polar equation $\theta = \frac{7\pi}{6}$. But this line passes through the origin, rather than the center of the hyperbola. The same inconsistency arises at the asymptote that occurs when $\theta \rightarrow (\frac{11\pi}{6})^±$.
There's a good video that outlines the process for sketching this curve (https://www.youtube.com/watch?v=y1l2R944W7s), and its creator posits that the actual asymptotes of the hyperbola have the same slopes as the lines $\theta = \frac{7\pi}{6}$, $\theta = \frac{11\pi}{6}$ ($\tan\frac{7\pi}{6}=\frac{1}{\sqrt{3}}$ and $\tan\frac{11\pi}{6}=-\frac{1}{\sqrt{3}}$, respectively), but they are shifted so as to pass through the center of the hyperbola (without offering any explicit mathematical justification for this). I can see that the nature/form of the hyperbola demands that this be true, but I'm still not totally comfortable with this behavior from a polar coordinate perspective. Can anybody explain the underlying mathematical reason that the equations for asymptotes in polar coordinates do not arise directly from the limit definition of the asymptotes as they do in Cartesian coordinates? In other words, is there a concrete explanation as to why an asymptote occurs at $x=a$ if $\lim_{x\to a^±}f(x)\rightarrow±\infty$, but $\lim_{\theta\to a^±}r(\theta)\rightarrow±\infty$ does not require that the asymptote be given by $\theta=a$ (beyond "it just doesn't")?
 A: In short, the apparent problem stems from trying to apply a rule of thumb for vertical asymptotes when working in rectangular Cartesian coordinates to arbitrary asymptotes in polar coordinates. If you plot ${6\over 1+2\sin x}$ on a Cartesian grid, you’ll find that the graph indeed has vertical asymptotes at $x=-\frac\pi6$ and $x=\frac{7\pi}6$, but then the graph no longer looks like a hyperbola.  
One way to understand why your reasoning led you astray for polar coordinates is to examine what happens to tangents to the curve as we approach an asymptote. On the Cartesian plane, for a differentiable function $f$, if $f(x)\to\infty$ as $x\to a$, then the slope of the tangent at $\left(x,f(x)\right)$ also tends to infinity, and the $x$-intercept of this tangent approaches $a$: loosely speaking, the limit of these tangent lines is the vertical asymptote $x=a$. Skipping the gory details of the calculation, the $x$-intercept of the tangent to $y={6\over 1+2\sin x}$ at $(x,y)$ is $x+\frac12\sec x+\tan x$, which equals $-\frac\pi6$ when $x=-\frac\pi6$ and $\frac{7\pi}6$ when $x=\frac{7\pi}6$. 
In polar coordinates, on the other hand, if $r\to\infty$ as $\theta\to\alpha$, the slope of the tangent approaches $\tan\alpha$, but we have no guarantee that these tangent lines tend toward the origin as you assumed. Indeed, the tangent to the hyperbola in your question at the point with polar coordinates $(r,\theta)$ has (again skipping the details of the derivation) the implicit Cartesian equation $x\cos\theta+y(2+\sin\theta)=6$. The square of the distance of this line from the origin is ${36\over 5+4\sin\theta}$, which approaches $12$, not $0$, as $\theta\to-\frac\pi6$ or $\theta\to\frac{7\pi}6$.
A: I had the same confusion, and I realised that it stemmed from not having a particularly robust idea of what an asymptote actually is. When I started thinking of it in terms of a function behaving increasingly like a given line in some part of its domain without ever actually intersecting that line, things became clearer. In polar coordinates, all lines parallel to a given ray, say $\theta=\alpha$, have the property that $r\to\infty$ as $\theta\to\alpha$ since the line, being parallel, does not intersect with $\theta=\alpha$. Thus, any function that behaves increasing like such a parallel line also has these properties. Hence, given that a function has those properties, all you can say is that the asymptote is parallel to the ray.
