Question

I need to convert 4/5-6costheta to a rectangular equation

Answer 1

I feel like I should start off with multiplying both sides by r but that gives me r in the numerator :/

Answer 2

so your equation is r=4/5 - 6 cos theta
and you want to convert it to rectangular form

Answer 3

yes

Answer 4

I've tried working it a few different ways and I always hit a dead end

Answer 5

in rectangular form you aren't going to get a function

Answer 6

I looked at a similar one, r= 6/7-8costheta and it gave the answer 15x^2+96x-49y^2+36=0 but I don't know how to get to that

Answer 7

those two aren't the same at all, graph them both and see

Answer 8

that's what my homework says.

Answer 9

https://www.desmos.com/calculator/gyz60mxqdi

Answer 10

you didn't enter it corrently

Answer 11

I guess that's my fault 6/(7-8costheta)

Answer 12

and the same with the other one?

Answer 13

Yes, sorry :(

Answer 14

so you can do it by analysis or by calculation. analysis is much easier, but won't get you very far if your teacher says to show your work

Answer 15

the only example he gave for this type of problem was one much more simple: r=3costheta. He multiplied both sides of the equation by r, then changed the left to x^2+y^2 and changed 3rcos(theta) to 3x and just solved from there

Answer 16

\[r=\frac{4}{5-6\cos \theta }\]
\[r^2=\frac{4r}{5-6\cos \theta }\]
\[x^2+y^2=\frac{4(x^2+y^2)}{5-6x }\]

Answer 17

correction: 4(sqrt(x^2+y^2)) in the numerators

Answer 18

okay, I was a little confused there. How did you change 6cos(theta) to 6x? I thought it had to be 6rcos(theta)?

Answer 19

sin theta = x/1, cos theta = y/1

Answer 20

*sin theta = y/1, cos theta = x/1

Answer 21

we can assume that r=1?

Answer 22

for the unit circle, yes

Answer 23

yes, but for the problem?

Answer 24

in my notes and the video I watched on youtube it says cos(theta)=x/r

Answer 25

so correction:
\[x^2+y^2=\frac{4r}{5-6\frac{ x }{ r } }\]
multiply by r/r
\[x^2+y^2=\frac{4r^2}{5r-6x}\]
multiply both sides by 5r-6x
\[(5r-6)\times(x^2+y^2)=4x^2+4y^2\]

Answer 26

hmmm, I went from there and divided each side by 4x^2+4y^2, factored out the 4 which made it cancel with the top. But again, it doesn't seem right

Answer 27

Hmm you guys are approaching this a little weird :O
I would probably do it like this:\[\Large r=\frac{4}{5-6\cos \theta}\]Multiply through by that denominator:\[\Large r(5-6\cos \theta)=4\]Distribute:\[\Large 5\color{orangered}{r}-6\color{royalblue}{r \cos \theta}=4\]

Use your identities:\[\Large \color{royalblue}{r \cos \theta=x}\]\[\Large \color{orangered}{r=x^2+y^2}\]

\[\Large 5\color{orangered}{(x^2+y^2)}-6\color{royalblue}{x}=4\]And then from there I guess you just need to complete the square on the x's to solve for y.

Answer 28

Yeahhhh, I didn't trust what he was doing and was waiting for him to admit that he didn't know either, haha

Answer 29

lol :)

Answer 30

but I thought r=(x^2+y^2)^(1/2)?

Answer 31

\[\Large \sin^2\theta+\cos^2\theta=1\]Multiply each side by r^2,\[\Large (\color{#DD4747 }{r \sin \theta})^2+(\color{royalblue}{r \cos \theta})^2=r^2\]From here we can plug in our identities:\[\Large \color{#DD4747 }{r \sin \theta=y}\]\[\Large \color{royalblue}{r \cos \theta =x}\]
\[\Large (\color{#DD4747 }{y})^2+(\color{royalblue}{x})^2=r^2\]

Answer 32

It's the form of a circle.
It's how we relate trig back to rectangular, with circle stuff :p

Answer 33

but we just had r, not r^2

Answer 34

Ohhhh we just had r, XDDD haha good call..
I made a boo boo didn't I?

Answer 35

:)

Answer 36

So that last line should be:\[\Large 5\color{orangered}{\sqrt{x^2+y^2}}-6\color{royalblue}{x}=4\]

Answer 37

Oh it wasn't a squared r.. doh!
Then maybe pete's method was on the right track :3

Answer 38

yeah, that's what I got when I continued his, yours was just a lot faster

Answer 39

Hmm so if you want to solve for y, you have to do some messy steps from here I guess.\[\Large 5\sqrt{x^2+y^2}\quad=\quad 6x+4\]Squaring:\[\Large 25x^2+25y^2=36x^2+48x+16\]And then you can probably solve for y now :O

Answer 40

ohhh, okay, I was making it a lot more messy by squaring the whole left side without moving the 6x to the right, haha

Answer 41

Ooo yah that could be a little tricky.

Answer 42

thank you :)

Answer 43

np \c:/