Question #02b85

1 Answer
Aug 4, 2015

Answer:

#x=1/8 y^2-2#.

Explanation:

One thing you can do is start by multiplying both sides of the equation #r=4/(1-cos(theta))# by #1-cos(theta)# to get #r-r cos(theta)=4#.

Next, rearrange this to get #r=4+r cos(theta)#.

Now square both sides to get #r^2=16+8r cos(theta)+r^2 cos^{2}(theta)#.

The reason this was a good idea is that you can now substitute rectangular coordinates #(x,y)# pretty quickly using the facts that #r^{2}=x^{2}+y^{2}# and #r cos(theta)=x# to get:

#x^2+y^2=16+8x+x^2#
#y^2=16+8x#.

Solving this equation for #x# as a function of #y# gives

#x=(1/8)(y^2-16)=1/8 y^2-2#.

The graph of #r=4/(1-cos(theta))#, as #theta# varies over the open interval #(0,2pi)#, is the sideways parabola shown below.

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