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i would like to know an easy way to find the asymptotes of the $$r = \frac1{1+2\cos \theta}.$$ this is for a precalculus audience, so it will be nice if calculus can be avoided altogether. i know that the asymptotes are parallel to the lines $\theta = \pm \pi/3.$ we also know that the $x$-intercepts are $1/3, 1.$

my audience don't know how to use the symmetry, most of them won't even know that the graph is a hyperbola. can we find the answer without using the symmetry of the graph.

abel
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    Well, the function is undefined when $\cos\theta =-\frac{1}{2}$, and, as $\theta$ approaches these, you get the asymptotes. – Thomas Andrews May 07 '15 at 15:57
  • @ThomasAndrews, the direction of the asymptotes is ok but we need to know one point on each so that we can sketch the graph. i can add this information if that helps. – abel May 07 '15 at 15:59
  • You could always convert the polar equation back into Cartesian. It is after all a hyperbola. From there you can easily derive the asymptotes. Cartesian equations of the hyperbola is also Pre-Calculus material I think? – imranfat May 07 '15 at 16:15
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    @imranfat, we are doing quick sketching of the polar graphs from the graphs of $r$ as a function of $\theta.$ we have done the easy ones $r =1+\sin \theta, \cdots.$ to sketch this graph right, you need more than the direction of the asymptotes. yes, we can convert it to cartesian and find the asymptotes. but i thought i would ask here and see if anyone has a different idea. – abel May 07 '15 at 16:26
  • Ah I see +1. Well, if there is an easy way to find the equations of the asymptotes from the polar equation, I also would like to know... – imranfat May 07 '15 at 17:37

2 Answers2

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Change basis to $\,\vec u_{2\pi/3}$ (the unit vector with polar angle $\frac{2\pi}3$) and the orthogonal vector $\vec v_{2\pi/3}$. An asymptote with polar angle $\frac{2\pi}3$ then has equation $Y=r\sin\bigl(\theta-\frac{2\pi}3\bigr)=l$. Hence you have to determine: $$\lim_{\theta\to\tfrac{2\pi}3}r(\theta)\sin\Bigl(\theta-\frac{2\pi}3\Bigr). $$ With some trigonometry ($\cos p-\cos q= \dots$), you should find a limit equal to $\frac 1{\sqrt 3}$.

Bernard
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here is what i came up with. we know that the graph of $r = \frac{1}{1+2\cos \theta}$ as a function of $\theta$ has vertical asymptotes at $\theta = 2\pi/3, 4\pi/3.$

we will show that for $\theta$ near $2\pi /3$ the polar graph is the graph of the slant asymptote of the hyperbola that cuts the $x$-axis at $\frac23.$

let $ \theta = 2\pi/3 + h.$ then $$\begin{align} r &= \frac1{1 + 2\cos(2\pi/3+h)}\\ &=\frac1{1+2\left(\cos(2\pi/3)\cos h - \sin(2\pi/3)\sin h \right)} \\ &=-\frac1{\sqrt3\sin h + \cdots} \\ &=-\frac1{\sqrt 3 \sin (t-2\pi/3) + \cdots}\end{align}$$

therefore one of the asymptotes is $$r =-\frac1{\sqrt3\sin (t-2\pi/3)},\, r(0) = \frac23 $$ as claimed.

in the same way we find that the second asymptote is the line $$r = \frac1{\sqrt3 \sin(t - 4\pi/3)}, \, r(0) = \frac23$$ and we see that the two asymptotes meet at the $x$-intercept $(\frac23, 0).$

abel
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