# Polar Form of Equation

• Apr 11th 2006, 05:47 PM
c_323_h
Polar 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 PM
CaptainBlack
Quote:

Originally Posted by c_323_h
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?

Introduce the polar coordinates $\displaystyle \{r,\theta \}$, such that:

$\displaystyle x=r \cos(\theta)$
$\displaystyle y=r \sin(\theta)$,

then substitute these into your equations.

RonL
• Apr 12th 2006, 03:46 AM
c_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 AM
topsquark
Quote:

Originally Posted by c_323_h
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?

a)
$\displaystyle 2(rcos\theta)-3(rsin\theta)=6$
$\displaystyle r(2cos\theta-3sin\theta)=6$
$\displaystyle r=\frac{6}{2cos\theta-3sin\theta}$

-Dan