How do you convert $r csc θ = 8$ $r csc theta = 8$ to rectangular form?

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Answer 1 · 2016-05-13T03:55:52

$x^{2} + y^{2} - 8 y = 0$ $x^2+y^2-8y=0$

Relation between polar form $(r, θ)$ $(r,theta)$ and rectangular form $(x, y)$ $(x,y)$ is given by

$r cos θ = x$ $rcostheta=x$ , $r sin θ = y$ $rsintheta=y$ , $r^{2} = a^{2} + b^{2}$ $r^2=a^2+b^2$ and $tan θ = \frac{b}{a}$ $tantheta=b/a$

Hence $r csc θ = 8$ $rcsctheta=8$ can be written as

$\frac{r}{sin θ} = 8$ $r/sintheta=8$ or

$\frac{r^{2}}{r sin θ} = 8$ $r^2/(rsintheta)=8$ or

$\frac{x^{2} + y^{2}}{y} = 8$ $(x^2+y^2)/y=8$ or $x^{2} + y^{2} = 8 y$ $x^2+y^2=8y$ or

$x^{2} + y^{2} - 8 y = 0$ $x^2+y^2-8y=0$

This is the equation of a circle.

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

How do you convert r csc theta = 8rcscθ=8r csc theta = 8 to rectangular form?