# How can I find the points of intersection between the curves $r=1+\sin\theta$ and $r=1-\sin\theta$?

Find the points of intersection for the curve $r=a(1+\sin\theta)$ and $r=a(1-\sin\theta)$

My book says the answer is $(0,0),(a,0),(a,\pi)$.

However I calculated $(a,0),(a,\pi),(a,2\pi)$.

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Your calculation is correct: you just forgot the case $a=0$. – awllower Apr 9 '13 at 4:23

Two of your points, $(a,0)$ and $(a,2\pi)$, are the same.

It is easy to miss the solution $(0,0)$, since $(0,\theta)$ is the origin whatever $\theta$ is: the origin has many polar addresses.

So $(0,0)$ is on one of the curves in the disguise $(0,\pi/2)$, and on the other in the disguise $(0,3\pi/2)$.

To spot this sort of thing, one has to examine algebraically the possibility that $r=0$ for some $\theta$. Or make a sketch.

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Suppose that $(r_0,\theta_0)$ is a solution to both $r=a(1+\sin(\theta))$ and $r=a(1-\sin(\theta))$.

That is, suppose $$r_0=a(1+\sin(\theta_0))\quad\text{and}\quad r_0=a(1-\sin(\theta_0)).$$ If $a=0$, then we must have $r_0=0$ and $\theta_0$ can be anything (but all of these points are the same, namely, they all describe the origin; the origin doesn't have a well-defined angle).

Now suppose that $a\neq 0$. Then we can divide both sides by $a$ to see that \begin{align*} a(1+\sin(\theta_0))=a(1-\sin(\theta_0)) &\iff 1+\sin(\theta_0)=1-\sin(\theta_0)\\ &\iff 2\sin(\theta_0)=0\\ &\iff \sin(\theta_0)=0\\ &\iff \theta_0\in\{0,\pi\} \end{align*} so that the solutions are $$(a(1+\sin(0)),0)=(a,0)\qquad\text{ and }\qquad (a(1+\sin(\pi)),\pi)=(a,\pi).$$

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makes sense thank you! – lawlipop Apr 9 '13 at 4:27
Glad I could help! – Zev Chonoles Apr 9 '13 at 4:28

For intersection, $1-\sin\theta=1+\sin\theta\implies \sin\theta=0\implies \theta=n\pi$ where $n$ is any integer

$\implies r=a(1\pm \sin n\pi)=a$

As the period of $\sin\theta,\cos\theta$ is $2\pi$

So, the in-congruent values of $\theta$ are $0,\pi$

Alternatively,

If $r=0,$ from the first equation $\sin\theta=-1$ and $\sin\theta=1$ from the second

$\implies r\ne0$

$r=a(1+\sin\theta)\implies r^2= ar+ ar\sin\theta\implies x^2+y^2-ar=ay$

Similarly, $r=a(1-\sin\theta)\implies x^2+y^2-ar=-ay$

$\implies ay=-ay\implies y=0\implies r=|x|, x^2-a|x|=0\implies |x|=a$ as $x=0$ would imply $r=0$

So, $x=\pm a\implies r=a,\theta= \arctan \frac{0}{\pm a}=n\pi$ whose in-congruent values are $0,\pi$

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