- Aug 7, 2001
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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?
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: 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.
Originally posted by: Deslocke
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: 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.
Originally posted by: Deslocke
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: 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.
