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)#