# Did I answer these questions correctly?

• Oct 3rd 2012, 10:43 PM
TWN
Did I answer these questions correctly?
1.) Convert the rectangular equation x^2 + y^2 - 2ax = 0 to its polar equivalent.

My answer is r = sqrt(2ax)

Not confident that one's right.

2.) Convert the rectangular equation xy = 4 to its polar equivalent.

My answer is (r^2)sin(theta)cos(theta) = 4

Not sure if that's a complete answer.

3.) Convert the rectangular equation (x^2 + y^2)^2 - 9(x^2 + y^2) to its polar equivalent.

The way I got that answer seemed waaaay too easy.

4.) Convert polar equation r = cot(theta)csc(theta) to its rectangular equivalent.

My answer is x = y^2

I'm pretty confident about that one.

Please look over what I've done and let me know if any of my answers are wrong. If one is wrong, please explain how to reach the correct answer.

• Oct 3rd 2012, 10:55 PM
Prove It
Re: Did I answer these questions correctly?
Quote:

Originally Posted by TWN
1.) Convert the rectangular equation x^2 + y^2 - 2ax = 0 to its polar equivalent.

My answer is r = sqrt(2ax)

Not confident that one's right.

2.) Convert the rectangular equation xy = 4 to its polar equivalent.

My answer is (r^2)sin(theta)cos(theta) = 4

Not sure if that's a complete answer.

3.) Convert the rectangular equation (x^2 + y^2)^2 - 9(x^2 + y^2) to its polar equivalent.

The way I got that answer seemed waaaay too easy.

4.) Convert polar equation r = cot(theta)csc(theta) to its rectangular equivalent.

My answer is x = y^2

I'm pretty confident about that one.

Please look over what I've done and let me know if any of my answers are wrong. If one is wrong, please explain how to reach the correct answer.

I disagree with Q1. Your final answer should NOT have any x's or y's left in it.

Q2 seems fine.

Q3 is not an equation.

Q4 seems fine.
• Oct 3rd 2012, 11:01 PM
TWN
Re: Did I answer these questions correctly?
What is the correct way to approach 1 and 3, please?
• Oct 3rd 2012, 11:05 PM
TWN
Re: Did I answer these questions correctly?
Oh wow, 3 is way easier than I first made it out to be. r = 3, correct?

Edit: And for #1, is r = 2a(cos(theta)) correct?
• Oct 3rd 2012, 11:14 PM
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Re: Did I answer these questions correctly?
Quote:

Originally Posted by TWN
Oh wow, 3 is way easier than I first made it out to be. r = 3, correct?

Edit: And for #1, is r = 2a(cos(theta)) correct?

I still disagree with 3. Are you sure you haven't made a typo? I agree with your new answer to 1.
• Oct 3rd 2012, 11:28 PM
TWN
Re: Did I answer these questions correctly?
Ah, I see what you're getting at now. The rectangular equation is (x^2 + y^2)^2 - 9(x^2 + y^2) = 0

• Oct 3rd 2012, 11:33 PM
Prove It
Re: Did I answer these questions correctly?
Quote:

Originally Posted by TWN
Ah, I see what you're getting at now. The rectangular equation is (x^2 + y^2)^2 - 9(x^2 + y^2) = 0

OK, so what do you get for the polar form of this equation?
• Oct 4th 2012, 12:08 AM
TWN
Re: Did I answer these questions correctly?
r = 3

Is that not right?

Here are my steps:

(x^2 + y^2)^2 - 9(x^2 + y^2) = 0

r^4 - 9r^2 = 0

r^4 = 9r^2

r^2 = 9

r = 3
• Oct 4th 2012, 12:14 AM
Prove It
Re: Did I answer these questions correctly?
Quote:

Originally Posted by TWN
r = 3

Is that not right?

Here are my steps:

(x^2 + y^2)^2 - 9(x^2 + y^2) = 0

r^4 - 9r^2 = 0

r^4 = 9r^2

r^2 = 9

r = 3

There are more solutions than that. The first fundamental rule is that you may not divide by 0. You need to factorise the equation then set each factor equal to 0.