Polar Form of Equation

**c_323_h** · April 11th 2006, 05:47 PM

Find the polar form of

a) $2x-3y=6$
b) $y^2=4x$

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?

**CaptainBlack** · April 11th 2006, 08:35 PM

Originally Posted by c_323_h

Find the polar form of

a) $2x-3y=6$
b) $y^2=4x$

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 $\{r,\theta \}$ , such that:

$<br /> x=r \cos(\theta)$
$<br /> y=r \sin(\theta)$ ,

then substitute these into your equations.

RonL

**c_323_h** · April 12th 2006, 03:46 AM

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

**topsquark** · April 13th 2006, 03:23 AM

Originally Posted by c_323_h

Find the polar form of

a) $2x-3y=6$
b) $y^2=4x$

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

-Dan

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