Question

convert to an equation in rectangular coords: r=1/(2-cos(theta))

Answer 1

cosθ = x/r

r = 1/(2 - x/r)

Answer 2

do you know what a "perfect trinomial square" is?

Answer 3

yes

Answer 4

But what does that have to do with this?

Answer 5

hmm one sec

Answer 6

\(\bf {\color{blue}{ r^2=x^2+y^2\implies r=\sqrt{x^2+y^2}\qquad x=rcos(\theta)}}\\
--------------------------\\
r=\cfrac{1}{2-cos(\theta)}\implies r[2-cos(\theta)]=1\implies 2r-2cos(\theta)=1\\ \quad \\
2(\sqrt{x^2+y^2})-x=1\implies 2(\sqrt{x^2+y^2})=x+1\\ \quad \\
\textit{squaring both sides}\\ \quad \\
2^2(\sqrt{x^2+y^2})^2=(x+1)^2\implies 4(x^2+y^2)=x^2+2x+1\\ \quad \\
4x^2+4y^2=x^2+2x+1\quad \textit{now grouping a few}\\ \quad \\
3x^2+2x+4y^2=1\implies (3x^2+2x)+4y^2=1\)

Answer 7

hmm I have a quick typo, dohh anyhow \(\bf{\color{blue}{ r^2=x^2+y^2\implies r=\sqrt{x^2+y^2}\qquad x=rcos(\theta)}}\\
--------------------------\\
r=\cfrac{1}{2-cos(\theta)}\implies r[2-cos(\theta)]=1\implies 2r-rcos(\theta)=1\\ \quad \\
2(\sqrt{x^2+y^2})-x=1\implies 2(\sqrt{x^2+y^2})=x+1\\ \quad \\
\textit{squaring both sides}\\ \quad \\
2^2(\sqrt{x^2+y^2})^2=(x+1)^2\implies 4(x^2+y^2)=x^2+2x+1\\ \quad \\
4x^2+4y^2=x^2+2x+1\quad \textit{now grouping a few}\\ \quad \\
3x^2+2x+4y^2=1\implies (3x^2+2x)+4y^2=1\)

Answer 8

\(\bf (3x^2+2x)+4y^2=1\implies 3\left(x^2+\frac{2}{3}x+{\color{red}{ \square }}^2\right)+4(y-0)^2=1\)

so... we would just need to make the 1st group, with the "x", a perfect square trinomial,
thus we'd need a value that would give us the middle term

Answer 9

if you notice the equation a bit, it'd look like an "ellipse" equation

Answer 10

we know that the middle term will be , 2 times the square of the left and right terms
so, what we have there is \(\bf \frac{2}{3}x\) that means that the other term will be \(\bf 2\cdot x\cdot {\color{red}{ \square }}=\cfrac{2}{3}x\implies {\color{red}{ \square }}=
\cfrac{2\cancel{x}}{3\cdot 2\cdot \cancel{x}}\)

Answer 11

so... what do you think is our trinomial?

Answer 12

I'm still confused... why can't you stop at (3x2+2x)+4y2=1

Answer 13

well, I guess you can
but I assume you're expected to keep it in some "standard form", thus

Answer 14

anyhow, if you don't need it as an ellipse standard form, then use -> (3x2+2x)+4y2=1 <- and that'd work

Answer 15

Actually my prof did say something about completing the square. How did you get 2x/3⋅2⋅x?

Answer 16

I'm very confused by this idea of polar coordinates.