Question

I need help converting r = 1 / ((5COS(THETA) - 8SIN(THETA)) INTO an equivelent rectangular equation.

Answer 1

\[x =r \cos \theta\]\[y =r \sin \theta\] this should help solve for cos and sin and put in your equation

Answer 2

make the substitutions:
\[\frac{x}{8} = 5 \cos \theta\]
\[\frac{y}{5} = 8 \sin \theta\]
this makes
r = 40

Answer 3

did you figure it out?
\[40 = \frac{1}{\frac{x}{8} - \frac{y}{5}} = \frac{40}{5x-8y}\]
\[5x -8y = 1\]
\[y = \frac{5x-1}{8}\]

Answer 4

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Answer 5

n that case a simpler method of solving it would be just transforming the quotion
r(5cos(theta) - 8sin(theta) = 1 : distrubute the r
5rcos(theta) - 8rsin(theta) = 1 : x=rcos(theta), y = rsin(theta)
5x - 8y = 1 solving for y yields the the same result as yours y = (5x - 1) / 8

but I don't think thats the answer

Answer 6

the 2 equations are equivalent

http://www.wolframalpha.com/input/?i=r+%3D+1%2F%285cos%28theta%29+-8sin%28theta%29%29+from+theta+%3D+0+to+2pi

Answer 7

is 5x - 8y = 1 equal to (x/5) + (y/8) = 1 ?

Answer 8

no

Answer 9

the polar equation
\[r = \frac{1}{5\cos \theta -8 \sin \theta}\]
is equal to line
\[y = \frac{5x-1}{8}\]

Answer 10

okay, thank you!

Answer 11

yw