Rectangular to Polar Conversion

[formalistlion]

This is really Pre-calc, but there wasn't a forum for that. I need to convert this equation from rectangular form to polar form:

y^2 -8x-16=0

I know how to convert rectangular to polar, but this ONE problem is giving me a hard time. My gut tells me there is a trig identity involved, but I could be wrong.

Thanks!

 

[renegade05]

What does y equal in polar ? what does x equal ?

is y not just y=r*sin(theta) and x=r*cos(theta)?

plug these in and solve for r using the quadratic formula. Graph your original equation and graph your new polar equation. You should see they are the same. Also remember to get all the values we need theta to go from 0-2pi.

(Sorry, i would use latex but it is still down...)

 

[formalistlion]

What does y equal in polar ? what does x equal ?

is y not just y=r*sin(theta) and x=r*cos(theta)?

plug these in and solve for r using the quadratic formula. Graph your original equation and graph your new polar equation. You should see they are the same. Also remember to get all the values we need theta to go from 0-2pi.

(Sorry, i would use latex but it is still down...)

Yeah, I know that part. The part I can't get is the rest. So we have (r*sin(theta))^2-8(r*cos(theta))-16=0. Now what? How do I get r by itself?

 

[renegade05]

Yeah, I know that part. The part I can't get is the rest. So we have (r*sin(theta))^2-8(r*cos(theta))-16=0. Now what? How do I get r by itself?

notice you have the form Ar^2+Br+C=0, where A=sin(theta)^2, B=cos(theta), C=-16.

Use the quadratic formula and substitute these values in to solve for r.

Also, once used it will be handy to recall that sin(x)^2+cos(x)^2 = 1