Question

Convert the polar equation to rectangular form
r= 3sec theta/5sec theta+5

Answer 1

Do you know the transformations for x, y and r in terms of polar coordinates?

Answer 2

Yes, x=rcos(theta), y=rsin(theta) and r=sqareroot of x^2+y^2

Answer 3

Excellent so:
r = (3/cos(theta))/(5/cos(theta) + 5)

Answer 4

I left the r for a reason....

Answer 5

Okay, that makes sense so far.

Answer 6

now you can sub for cos(theta)....in terms of x and r

Answer 7

How exactly would you do that?

Answer 8

x=rcos(theta) therfore cos(theta) = x/r

Answer 9

therefore...

Answer 10

I'm not sure how to move from that step...

Answer 11

Try to rewrite this now
r = (3/cos(theta))/(5/cos(theta) + 5) with cos(theta) = x/r
( leave the r on the left for now)

Answer 12

So r= (3/(x/r))/(5/(x/r) + 5)?

Answer 13

yes..I would simplify the right hand side as much as possible then you can sub for r...
r = (x^2 + y^2)^1/2

Answer 14

Once you do that...it should all be just a function of x and y.

Answer 15

Could you also simplify in the beginning and start off with r=5cos(theta)+5/3cos(theta)?

Answer 16

hang on...

Answer 17

When I simply the original expression
r = (3/cos(theta))/(5/cos(theta) + 5)

I get
r = 3/(5 + 5cos(theta)

Answer 18

Oh okay. I still don't understand how you would substitute for r in the initial equation

Answer 19

r = (x^2 + y^2)^1/2

Answer 20

Would that be the square root ?

Answer 21

yes

Answer 22

What's the simplest the right side can get?

Answer 23

Can you simplify it down to 3cos(theta)+15/5cos(theta)

Answer 24

hang on

Answer 25

I am going to leave it to you to work it out.
If you want to post your final answer I would be happy to look at it.

Answer 26

One quick thing, how do I get rid of the square roots?

Answer 27

Square both sides.