How do you convert $x+2y-4=0$ to polar form?

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Answer 1 · 2016-05-04T16:15:32

$r=4cosalpha/cos(theta-alpha)$ where $alpha=tan^(-1)2$

If $(r,theta)$ is in polar form and $(x,y)$ in Cartesian form the relation between them is as follows:

$x=rcostheta$ , $y=rsintheta$ , $r^2=x^2+y^2$ and $tantheta=y/x$

Hence, $x+2y-4=0$ can be written as

$rcostheta+2rsintheta-4=0$ or

$r(costheta+2sintheta)=4$ or

$r=4/(costheta+2sintheta)$ .....(A)

Now let $tan^(-1)2=alpha$ or $2=tanalpha=sinalpha/cosalpha$

Hence (A) becomes $r=4/(costheta+(sinalpha/cosalpha)sintheta)$

or $r=4cosalpha/(costhetacosalpha+sinalphasintheta)$ or

$r=4cosalpha/cos(theta-alpha)$

How do you convert x+2y-4=0x+2y-4=0 to polar form?