Question

Convert the polar function r=cos(theta])-2sin(theta) to rectangular form.

What I got after simplifying everything is
(x^2+y^2)^3 = y^2 - 4x^2 would this answer be correct?

Answer 1

that is not what i get

Answer 2

putting \(\cos(\theta)=\frac{x}{r}=\frac{x}{x^2+y^2}\) and \(2\sin(\theta)=\frac{2y}{x^2+y^2}\)
i get
\[{x^2+y^2}=\frac{x-2y}{x^2+y^2}\] or \[(x^2+y^2)^4=x-2y\] but i am not positive this is right

Answer 3

no my answer is wrong
must have made a mistake somewhere

Answer 4

Here are the steps I got.
r =\[\frac{ y }{ r } - 2 \frac{ x }{ r }\]
multiply both sides by \[r^{2}\] to get
\[r ^{3} = y - 2x\] and we know r = \[\sqrt{x ^{2}+y ^{2}}\]square both sides to get ride of the square root to get \[(x^{2}+y ^{2})^{3} = y ^{2} - 4x ^{2}\]

Answer 5

i think starting with
\[r=\cos(\theta)-2\sin(\theta)\] or
\[r=\frac{x}{r}-\frac{2y}{r}\] multiply both sides by \(r\) gives
\[r^2=x-2y\] or
\[x^2+y^2=x-2y\] which is a circle

Answer 6

i made a silly mistake at the beginning, but i knew it was wrong because it had to be a circle

Answer 7

so my mistake would have just been multiplying by \[r^{2}\] instead of just r to cancel out denominators, and the silly mistake of mixing up x and y