Question

write the polar equation in rectangular form:\[r={1\over{9cos\theta-3sin\theta}}\]

Answer 1

do you know how to find polar form when r=acos +bsin ?

Answer 2

\[x = rcos \Theta, y =r\sin \Theta \] \]

Answer 3

yes @hartnn

Answer 4

so, here, you can multiply and divide by 9cos + 3 sin

Answer 5

so that denominator is of the form (a+b)(a-b)=..?

Answer 6

that makes sense

Answer 7

yeah, a constant in denominator and 9cos +3sin in numerator will brig the equation in the form of a cos +b sin

Answer 8

hmm...actually you don't get a constant in the denominator...

Answer 9

you have to multiply and divide by \(\sqrt{9^2+3^2}\) first.

Answer 10

wait that just gives me\[r={{9cos\theta+3sin\theta}\over{9cos^2\theta-3sin^2\theta}}\]

Answer 11

could i do this..

Answer 12

@hartnn To me there seems to be an easier way...

Answer 13

now if you multiply and divide by conjugate of denominator....
feel free to share... :)

Answer 14

\[\begin{array}{rcl}
r &=& \frac{1}{9\cos\theta - 3\sin\theta}\\
9r\cos\theta - 3r\sin\theta &=& 1 \\
9x-3y &=& 1
\end{array}
\]

Answer 15

\[r={1\over9cos\theta}-{1\over3sin\theta}\]\[r+{1\over3sin\theta}={1\over9cos\theta}\]\[9cos\theta\times r={3sin\theta}\]\[9r cos\theta\times r^2 = 3r sin \theta\]\[9x+x^2+y^2+=3y\]\[x^2+9x+{81\over4}+y^2-3y+{9\over4}={45\over2}\]\[(x+{9\over2})^2+(y-{3\over2})^2={45\over2}\]yes? :D

Answer 16

:O
so much simpler...how could i not think of this :P

Answer 17

wow...i feel stupid....................

Answer 18

i too :P

Answer 19

good thought @wio

Answer 20

that seems right and it is but im still unsure...idk why