rectangular form to polar form simplification

[scrilla103]

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]

(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]

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]

try the quaderatic formula like they said

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

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

[arletebacon]

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]

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]

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]

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]

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:

. . . . .

...or, which is the same thing:

. . . . .

[arletebacon]

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