So today's lecture was about polar coordinates, and we were taught about the concept up to limacons. I'd like to know why the graph of $r = a + b\cos \theta$ is exactly the same as the graph of $r = -a + b\cos \theta$ . I've tried substituting values for a and b but I still can't make sense of the results.
[Math] Why is the graph of $r = a + b\cos \theta$ the same whether a is positive or negative
calculuspolar coordinates
Related Solutions
I'd say that it's convenient to include it in the set of rose curves—it can be useful to think about $r=a\sin n\theta$ for noninteger values of $n$ (even starting at $0$), rather than just integers greater than $1$:
$$r = \cos \theta$$
$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$
$$x^2 + y^2 = x$$
$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$
$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$
This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius $1 \over 2$.
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Best Answer
$r=a+b \cos \theta$ and $r=-a+b \cos \theta$ look very much the same, but are offset by $\pi$ and negated.
In a normal plot, here with a=5,b=7, plotted from $\theta=-\pi .. \pi$ you see the difference:
+a:
-a:
If you plot the whole graph on polar plot, the graph is cyclic, so it does not matter where you start or end. The offset by $\pi$ rotates the plot by 180° and the negation of the radius rotates it again by 180°. So both effects chancel each other and the polar graphs look exactly the same. If you only plot a part of the graph you see the difference.
Here the full Polar Graph of both cases (a=5, b=7):
Now the +a case only plotted from $\theta = -\pi .. \pi/2$:
and the -a case only plotted from $\theta = -\pi .. \pi/2$: