How do you find the cartesian equation for r = 3 / (1 + cosT)r=31+cosT?

1 Answer
A. S. Adikesavan
May 14, 2016

y^2+6x-9=0y2+6x−9=0, representing a parabola, with vertex at (3/2, 0)(32,0) and axis along x-axis in the negative x-direction.

Explanation:

Reading T as the polar coordinate theta, cos theta=x/r and r^2=x^2+y^2θ,cosθ=xrandr2=x2+y2.

Now, the polar equation becomes 3-x = r = sqrt(x^2+y^2)3−x=r=√x2+y2.

Squaring and simplifying, y^2+6x-9=0y2+6x−9=0. The alternative standard

form is y^2=-6(x-3/2)y2=−6(x−32), from which the characteristics of this parabola as given in the answer. .

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