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math help (pre-calc)

thirtythree

Diamond Member
I was absent for a day so I'm slightly lost. Would someone be kind enough to explain how to change 9x^2-4y^2=36 into a polar equation?
 
x = r cos (theta)
y = r sin (theta)

plug those into your equation ...

The format of the original equation you gave is 9x^2 - 4y^2 = r^2 which means that r^2 = 36 and r = 6.

i get 9 cos^2 (theta) - 4 sin^2 (theta) = 1

i'm not sure, somebody check my work, i just did it in my head.
 
Originally posted by: jlarsson
x = r cos (theta)
y = r sin (theta)

plug that into your equation ... by the way, r = 6

i get 9 cos^2 (theta) + 4 sin^2 (theta) = 1

i'm not sure, somebody check my work, i just did it in my head.
Uhh .. can I ask how you got that? I can get r^2(9cos^2-4sin^2)=36 but how do you know r is 6? And is that plus in the middle supposed to be a minus?
 
Originally posted by: Deslocke
Originally posted by: jlarsson
x = r cos (theta)
y = r sin (theta)

plug that into your equation ... by the way, r = 6

i get 9 cos^2 (theta) + 4 sin^2 (theta) = 1

i'm not sure, somebody check my work, i just did it in my head.
Uhh .. can I ask how you got that? I can get r^2(9cos^2-4sin^2)=36 but how do you know r is 6? And is that plus in the middle supposed to be a minus?

yeah, its supposed to be a minus ... I edited my previous post.

 
Originally posted by: Deslocke
Originally posted by: jlarsson
x = r cos (theta)
y = r sin (theta)

plug that into your equation ... by the way, r = 6

i get 9 cos^2 (theta) + 4 sin^2 (theta) = 1

i'm not sure, somebody check my work, i just did it in my head.
Uhh .. can I ask how you got that? I can get r^2(9cos^2-4sin^2)=36 but how do you know r is 6? And is that plus in the middle supposed to be a minus?

You know that A*x^2+B*y^2=r^2
since you are given A=9 and B=4 then you know that r^2=36. .:r=6
 
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