How do you convert $r \cdot (csc θ) = \sqrt{11}$ $r*(csctheta) = sqrt(11)$ into rectangular form?

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Answer 1 · 2016-02-23T08:42:03

$x^{2} + y^{2} - \sqrt{11} y = 0$ $x^2+y^2-sqrt11y=0$

To convert $r*cscθ=sqrt11$ into rectangular form $(x,y)$ , we should use the relation between $(x,y)$ and $(r,theta)$ , which are

$r^2=x^2+y^2$ and $rcostheta=x$ and $rsintheta=y$
Note that $csctheta=1/sintheta=r/y$ .

Hence $r*cscθ=sqrt11$ can be written as

$r*r/y=sqrt11$ or $r^2=sqrt11*y$

$x^2+y^2-sqrt11y=0$

How do you convert r*(csctheta) = sqrt(11)r⋅(cscθ)=√11r*(csctheta) = sqrt(11) into rectangular form?