How do you convert the polar equation $r = \frac{3}{1 - sin θ}$ $r=3/(1-sintheta)$ into rectangular form?

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Answer 1 · 2017-02-02T07:40:19

The answer is $y = \frac{x^{2} - 9}{6}$ $y=(x^2-9)/6$

To convert from polar coordinates $(r, θ)$ $(r, theta)$ to rectangular coordinates $(x, y)$ $(x,y)$ , we use the following equations

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsin theta$

$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Therefore,

$r = \frac{3}{1 - sin θ}$ $r=3/(1-sintheta)$

$r (1 - sin θ) = 3$ $r(1-sintheta)=3$

$r - r sin θ = 3$ $r-rsintheta=3$

$r = 3 + r sin θ$ $r=3+rsintheta$

$\sqrt{x^{2} + y^{2}} = 3 + y$ $sqrt(x^2+y^2)=3+y$

$x^{2} + y^{2} = {(3 + y)}^{2} = 9 + 6 y + y^{2}$ $x^2+y^2=(3+y)^2=9+6y+y^2$

$x^{2} = 9 + 6 y$ $x^2=9+6y$

$6 y = x^{2} - 9$ $6y=x^2-9$

$y = \frac{x^{2} - 9}{6}$ $y=(x^2-9)/6$

How do you convert the polar equation r=3/(1-sintheta)r=31−sinθr=3/(1-sintheta) into rectangular form?