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]}

have a nice day ...from the Philippines !

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