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

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Answer 1 · 2016-06-20T09:24:45

$x^{2} + 4 y - 4 = 0$ $x^2+4y-4=0$

If polar coordinates $(r, θ)$ $(r,theta)$ are written as $(x, y)$ $(x,y)$ in Cartesian coordinates, then

$r cos θ = x$ $rcostheta=x$ , $r sin θ = y$ $rsintheta=y$ and $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

Hence, $r = \frac{2}{1 + sin θ}$ $r=2/(1+sintheta)$ can be written as

$r = \frac{2}{1 + \frac{y}{r}}$ $r=2/(1+y/r)$ or $r + y = 2$ $r+y=2$ or $r = 2 - y$ $r=2-y$ and on squaring

$r^{2} = y^{2} - 4 y + 4$ $r^2=y^2-4y+4$ i.e. $x^{2} + y^{2} = y^{2} - 4 y + 4$ $x^2+y^2=y^2-4y+4$ or

$x^{2} + 4 y - 4 = 0$ $x^2+4y-4=0$ , which is the equation of a parabola.

graph{x^2+4y-4=0 [-10, 10, -6.8, 3.2]}

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