How do you convert $r = \frac{2}{1 + sin (θ)}$ $r = 2 / (1+sin(theta))$ into rectangular form?

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Answer 1 · 2018-04-14T02:57:19

The rectangular equation is $y = - \frac{x^{2}}{4} + 1$ $y=-x^2/4+1$ .

$r = \frac{2}{1 + sin θ}$ $r=2/(1+sintheta)$

$r = \frac{2}{1 + sin θ} \cdot \frac{1 - sin θ}{1 - sin θ}$ $r=2/(1+sintheta)color(red)(*(1-sintheta)/(1-sintheta))$

$r = \frac{2 - 2 sin θ}{1 - {sin}^{2} θ}$ $r=(2-2sintheta)/(1-sin^2theta)$

$r = \frac{2 - 2 sin θ}{{cos}^{2} θ}$ $r=(2-2sintheta)/cos^2theta$

$r^{2} = \frac{2 r - 2 r sin θ}{{cos}^{2} θ}$ $r^2=(2r-2rsintheta)/cos^2theta$

$r^{2} {cos}^{2} θ = 2 r - 2 r sin θ$ $r^2cos^2theta=2r-2rsintheta$

${(r cos θ)}^{2} = 2 r - 2 r sin θ$ $(rcostheta)^2=2r-2rsintheta$

Using the substitutions $y = r sin θ$ $y=rsintheta$ and $x = r cos θ$ $x=rcostheta$ and $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ :

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

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

$\frac{x^{4}}{4} + y x^{2} + y^{2} = x^{2} + y^{2}$ $x^4/4+yx^2+y^2=x^2+y^2$

$\frac{x^{4}}{4} + y x^{2} = x^{2}$ $x^4/4+yx^2=x^2$

$x^{2} (\frac{x^{2}}{4} + y) = x^{2}$ $x^2(x^2/4+y)=x^2$

$\frac{x^{2}}{4} + y = 1$ $x^2/4+y=1$

$y = - \frac{x^{2}}{4} + 1$ $y=-x^2/4+1$

That's the equation; it's a parabola. Here's what it looks like:

graph{-x^2/4+1 [-10, 10, -5, 5]}

How do you convert r = 2 / (1+sin(theta))r=21+sin(θ)r = 2 / (1+sin(theta)) into rectangular form?