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

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Answer 1 · 2018-04-14T02:57:19

The rectangular equation is #y=-x^2/4+1#.

#r=2/(1+sintheta)#

#r=2/(1+sintheta)color(red)(*(1-sintheta)/(1-sintheta))#

#r=(2-2sintheta)/(1-sin^2theta)#

#r=(2-2sintheta)/cos^2theta#

#r^2=(2r-2rsintheta)/cos^2theta#

#r^2cos^2theta=2r-2rsintheta#

#(rcostheta)^2=2r-2rsintheta#

Using the substitutions #y=rsintheta# and #x=rcostheta# and #r=sqrt(x^2+y^2)#:

#x^2=2sqrt(x^2+y^2)-2y#

#x^2/2+y=sqrt(x^2+y^2)#

#x^4/4+yx^2+y^2=x^2+y^2#

#x^4/4+yx^2=x^2#

#x^2(x^2/4+y)=x^2#

#x^2/4+y=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]}