How do you convert #r = 4 / (1 - cos theta)# to rectangular form?

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Answer 1 · 2018-02-06T11:51:02

#y^2-8x-16=0#

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From the diagram we can see that point P has polar coordinates
#( r , theta )# and Cartesian coordinates #(x,y)#.

And #color(white)(88)x=rcos(theta)# , #y = rsin(theta)#

#(x,y) -> (rcos(theta), rsin(theta))#

Also:

By Pythagoras' Theorem :

#r^2=(rcostheta)^2+(rsintheta)^2#

Since:

#x=rcos(theta)# and #y = rsin(theta)#

Then:

#r^2=x^2+y^2# #:. r=sqrt(x^2+y^2)#

Using these ideas:

#r=4/(1-cos(theta))#

Substituting:

#sqrt(x^2+y^2)=4/(1-cos(theta))#

#cos(theta)=x/r#

#sqrt(x^2+y^2)=4/(1-x/r)#

#sqrt(x^2+y^2)=4/(1-x/(sqrt(x^2+y^2))#

Multiply by #(1-x/(sqrt(x^2+y^2)))#

#sqrt(x^2+y^2)-(xsqrt(x^2+y^2))/(sqrt(x^2+y^2))=(4(1-x/(sqrt(x^2+y^2))))/(1-x/(sqrt(x^2+y^2))#

#sqrt(x^2+y^2)-(xcancel(sqrt(x^2+y^2)))/(cancel(sqrt(x^2+y^2)))=(4(cancel(1-x/(sqrt(x^2+y^2)))))/((cancel(1-x/((sqrt(x^2+y^2))))))#

#sqrt(x^2+y^2)-x=4#

#sqrt(x^2+y^2)=4+x#

Squaring:

#x^2+y^2=x^2+8x+16#

#y^2-8x-16=0#