How do you convert #r = 8 sin θ# to rectangular form?

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Answer 1 · 2016-02-15T09:30:36

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

You can convert #r=8 sin theta# to rectangular form by using the following

#r=sqrt(x^2+y^2)# and #sin theta=y/sqrt(x^2+y^2)#

replace everything in the given

#r=8 sin theta#

#sqrt(x^2+y^2)=8*y/sqrt(x^2+y^2)#

simplify at this point by multiplying both sides by #sqrt(x^2+y^2)#

#sqrt(x^2+y^2)*sqrt(x^2+y^2)=sqrt(x^2+y^2)*8*y/sqrt(x^2+y^2)#

#sqrt(x^2+y^2)*sqrt(x^2+y^2)=cancelsqrt(x^2+y^2)*8*y/cancelsqrt(x^2+y^2)#

#(sqrt(x^2+y^2))^2=8*y#

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

#x^2+y^2-8y=0" " " "#the equivalent rectangular form

graph{x^2+y^2-8y=0[-20,20,-10,10]}

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