## rectangular form to polar form simplification

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scrilla103
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### rectangular form to polar form simplification

Equation given:
(y^2) - (8x) - (16) = 0
Convert to polar form and simplify.
{ I tried multiple routes to no avail. }
Please show steps with brief explanations, thanks

stapel_eliz
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(y^2) - (8x) - (16) = 0
Convert to polar form and simplify.
{ I tried multiple routes to no avail. }
Unfortunately, it is not possible to help you find any errors in work we cannot see. So please reply showing one of the routes you tried, starting from the standard substitution and proceeding through using the Quadratic Formula to solve for "r=". Thank you!

scrilla103
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### Re: rectangular form to polar form simplification

So here is the basic process I tried:
(y^2) - (8x) - (16) = 0
**substitute y with rsint and x with rcost**
(r^2)[(sint)^2] - (8rcost) - (16) = 0 {In words: "(r squared sine squared theta) minus (eight r cosine theta) minus (sixteen) equals (zero)}

This is where I get stuck. I cannot seem to simplify this equation to have "r" alone on one side... there is always an "r" on the other side.

Any ideas?
Thanks.

little_dragon
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### Re: rectangular form to polar form simplification

try the quaderatic formula like they said

sin^2(@) r^2 - 8cos(@) r - 16 = 0

so a=sin^2(@), b=-8cos(@), c=-16

arletebacon
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### Re: rectangular form to polar form simplification

I have the same question

Y^2-8x-16=0. The answer listed is r= 16/(-7sin(theta))

can someone explain how to get there!!!!

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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### Re: rectangular form to polar form simplification

I have the same question

Y^2-8x-16=0. The answer listed is r= 16/(-7sin(theta))

can someone explain how to get there!!!!
Yes: read the previous replies, which provide an explanation of "how to get there".

arletebacon
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### Re: rectangular form to polar form simplification

I have read the posts and tried all those suggestions.. but I don't get the answer that was provided by my course.

i don't know how to get the r= 16/(-7sin(theta)) as an answer.

arletebacon
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### Re: rectangular form to polar form simplification

a=sin^2(theta)
b=-8cos(theta)
c=-16

(-b+-sqrt(b2-4ac))/2a

8cos(theta) +- sqrt( (-8cos(theta)^2) - (4) (sin^2(theta)) (-16) ) / 2(sin^2(theta)

8cos(theta) + - sqrt ( 64cos^2(theta) +64 sin^2(theta) ) / 2 sin^2(theta)

( 8cos (theta) +- 8 ) / 2 sin^2(theta)

4(cos(theta) +- 4) / sin^2(theta)

4( cos(theta) _+ 1) / 1- cos^2(theta)

now I just went with the 4costheta + 1 version.. there is still the 4costheta - 1

4 (cos(theta) +1) / (1-cos(theta)) (1+cos(theta))

4 / 1-cos(theta) ..............

I don't know how to get from there to here r= 16/(-7sin(theta))

stapel_eliz
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I hadn't checked the work all the way to the answer. But the solution key is wrong; you're right. The polar form is:

. . . . .$r\, =\, \frac{4}{1\, -\, \cos\left(\theta\right)}$

...or, which is the same thing:

. . . . .$4\, =\, r\,-\, r\cos\left(\theta\right)$

arletebacon
Posts: 5
Joined: Thu Oct 06, 2011 4:30 am
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### Re: rectangular form to polar form simplification

thank you..!! I spent tons of hours on that... appreciate your help!!