Converting an equation from polar to recangular

Question

I'm having trouble converting the following equation from polar to rectangular:

$$ r = \frac{4}{3+2\cos(\theta)} $$

Every method I try still leaves either a $\theta$ or an $r$ in there.

Can I get some help?

Answer 1

We have the relationships ${r^2} = {x^2} + {y^2}$, $x = r \cos(\theta)$, and $y = r \sin(\theta)$ that relate polar and rectangular coordinates. If we start with your problem: $$r = \frac{4}{3+2\cos\theta}$$ multiplying both sides by the denominator and distributing gives $$3r + 2r \cos\theta=4.$$ Using the equations above, we see that $r=\sqrt{x^2+y^2}$ and $r \cos \theta = x$, so we can rewrite the above equation as $$3\sqrt{x^2+y^2}+2x=4.$$ Subtracting $2x$ from both sides and squaring both sides gives $$9(x^2+y^2)=(4-2x)^2=16-16x+4x^2.$$ Rearranging this a bit, we get $$\eqalign{9(x^2+y^2) &= 16-16x+4x^2 \cr 9x^2 + 9y^2 &= 16 - 16x + 4x^2 \cr 5x^2 + 16x + 9y^2 &= 16}$$ Now we need complete the square on $5x^2+16x$: $$\eqalign{5x^2 + 16x + 9y^2 &= 16 \cr 5(x^2+\frac{16}{5}x) + 9y^2 &= 16 \cr 5(x^2 + \frac{16}{5}x + \frac{64}{25})+9y^2 &= 16 + 5\cdot\frac{64}{25} \cr 5(x+\frac{8}{5})^2+9y^2 &= \frac{144}{5} \cr \frac{25(x+\frac{8}{5})^2}{144}+\frac{45 y^2}{144} &= 1 \cr \frac{(x+\frac{8}{5})^2}{144/25}+\frac{y^2}{144/45} &= 1 }$$ This is standard rectangular form for an ellipse with center $(-\frac{8}{5},0)$, $x$-radius $\frac{12}{5}$, and $y$-radius $\frac{12}{\sqrt{45}}$.

Answer 2

$$3r=4-2r\cos\theta\implies 9r^2=(4-2r\cos\theta)^2$$ So we get $$9(x^2+y^2)=(4-2x)^2$$ which will simplify to a conic. In general, polar equations of the form $$r=\frac{a}{1+e\cos\theta}$$ are always conics, it would be helpful to study the various cases because they come up quite a lot (e.g., in most derivations of Kepler's Laws from Newton's law of gravitation).