Question

Find an equation in x and y that has the same graph as the polar equation: r = 8 sin theta - 2 cos theta

Answer 1

\(\bf r = 8sin(\theta)-2cos(\theta)\\
\textit{let's multiply both sides by "r"}\\
r^2 = r(8sin(\theta)-2cos(\theta))\)

so, what do you think?

Answer 2

I don't know why I didn't think of that. Maybe I've been doing too much math today :P

Answer 3

heheh

Answer 4

so multiply and expand the right-hand side then, what do you get?

Answer 5

r^2 = 8 r sin theta - 2 r cos theta
r^2 = 8y - 2x

Answer 6

yes, and \(\bf x^2+y^2 = r^2\)

Answer 7

so we get \[x^2 + y ^2 = 8y-2x\] but what then?

Answer 8

then you do do a completion of the square
\(\bf x^2 + y ^2 = 8y-2x \implies x^2 + y ^2-8y+2x =0\\
x^2+2x+y^2-8y=0\\ \implies (x^2+2(1)x+\square?) +(y^2-2(4)y+\square?)=0\)

Answer 9

after which you'd end up with an circle :)

Answer 10

I forgot how to complete the square...

Answer 11

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

Answer 12

\[(x^2 + 2x +1) + (y^2 -8y +16) =0\] is that it?

Answer 13

to complete the square you add stuff that will give you a perfect square trinomial
just don't forget to subtract the same amount, because all we're doing is borrowing from Zero (0)
so +25 -25 = 0
+1,000 - 1,000 = 0

Answer 14

yes

Answer 15

I had to look it up lol and wouldn't that mean I had to also subtract 17 from that side or at least add it to 0 on the other side?

Answer 16

\(\bf (x^2 + 2x +1) + (y^2 -8y +16) =0\\
\textit{we added }1^2+4^2 \textit{ then we also subtract } -1^2-4^2\\
(x^2 + 2x +1) + (y^2 -8y +16)-17 =0\\ \implies (x^2 + 2x +1) + (y^2 -8y +16) =17\)

Answer 17

Thats what I thought..

Answer 18

\(\bf (x+1)^2+(y-4)^2 = 17 \) and that's the circle :)

Answer 19

Is there no other way to simplify it?

Answer 20

don't think so

Answer 21

THANK YOU!!!

Answer 22

yw