Find the polar form of

a) $\displaystyle 2x-3y=6$

b) $\displaystyle y^2=4x$

I can manipulate the equation all I want but I don't know when I've got the polar form. Is there a standard form or something?

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- Apr 11th 2006, 05:47 PMc_323_hPolar Form of Equation
Find the polar form of

a) $\displaystyle 2x-3y=6$

b) $\displaystyle y^2=4x$

I can manipulate the equation all I want but I don't know when I've got the polar form. Is there a standard form or something? - Apr 11th 2006, 08:35 PMCaptainBlackQuote:

Originally Posted by**c_323_h**

$\displaystyle

x=r \cos(\theta)$

$\displaystyle

y=r \sin(\theta)$,

then substitute these into your equations.

RonL - Apr 12th 2006, 03:46 AMc_323_h
but after i do that i can still manipulate the equation many times. how do i know which equation to write? do i keep going until i can no longer simplify?

this is my answer for the first one.

r=6/2cosx-3sinx

and for the second one

r^2(sin^2)-4rcos(x)=0 - Apr 13th 2006, 03:23 AMtopsquarkQuote:

Originally Posted by**c_323_h**

$\displaystyle 2(rcos\theta)-3(rsin\theta)=6$

$\displaystyle r(2cos\theta-3sin\theta)=6$

$\displaystyle r=\frac{6}{2cos\theta-3sin\theta}$

-Dan