# How do i solve this trig problem - Polar equations.

• Jan 12th 2011, 06:40 PM
Jackschmerold
How do i solve this trig problem - Polar equations.
Here is the problem that I cannot figure out how to do. Please explain.

r = (5csc(θ))/(6csc(θ) + 3)
possibilities:
a) (36x^2) + (27y^2) + (30x) -25 = 0
b) (27x^2) + (36y^2) +(30x) -25 = 0
c) (36x^2) + (27y^2) + (30y) -25 = 0
d) none of the above

Thankyou so much for any help
• Jan 12th 2011, 06:52 PM
Chris L T521
Quote:

Originally Posted by Jackschmerold
Here is the problem that I cannot figure out how to do. Please explain.

r = (5csc(θ))/(6csc(θ) + 3)
possibilities:
a) (36x^2) + (27y^2) + (30x) -25 = 0
b) (27x^2) + (36y^2) +(30x) -25 = 0
c) (36x^2) + (27y^2) + (30y) -25 = 0
d) none of the above

Thankyou so much for any help

What have you done so far? (Is this from a homework assignment by any chance?)
• Jan 12th 2011, 07:00 PM
Jackschmerold
Yes it is a hw problem.
So far:
6rcsc(θ) + 3r = 5cscθ
3r = 5cscθ - 6rcscθ
(3r)^2 = (5cscθ)^2 - (6rcscθ)^2
3x^2 + 3y^2 = ?
I dont know where to go from there
• Jan 12th 2011, 07:17 PM
Chris L T521
Quote:

Originally Posted by Jackschmerold
Yes it is a hw problem.
So far:
6rcsc(θ) + 3r = 5cscθ
3r = 5cscθ - 6rcscθ
(3r)^2 = (5cscθ)^2 - (6rcscθ)^2
3x^2 + 3y^2 = ?
I dont know where to go from there

Here's a suggestion. When you're here: $6r\csc\theta+3r=5\csc\theta$, multiply both sides by $\sin\theta$ to get $6r+3r\sin\theta=5$.

Now make the conversion to rectangular coordinates. Can you take from here?
• Jan 12th 2011, 07:17 PM
Chris L T521
Quote:

Originally Posted by Jackschmerold
Yes it is a hw problem.
So far:
6rcsc(θ) + 3r = 5cscθ
3r = 5cscθ - 6rcscθ
(3r)^2 = (5cscθ)^2 - (6rcscθ)^2
3x^2 + 3y^2 = ?
I dont know where to go from there

Here's a suggestion. When you're here: $6r\csc\theta+3r=5\csc\theta$, multiply both sides by $\sin\theta$ to get $6r+3r\sin\theta=5$.

Now make the conversion to rectangular coordinates. Can you take from here?