How do you express the polar equation #r^2 = 16cos theta# in cartesian form?

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Answer 1 · 2017-10-23T21:02:29

#(x^2+y^2)^(3/2) - 16x = 0#

To convert from polar to rectangular coordinates we can use:

#x = r cos theta#

#y = r sin theta#

and the consequence:

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

So, given:

#r^2 = 16cos theta#

we can multiply both sides by #r# to get:

#r^3 = 16 r cos theta#

Then use some of our formulae to rewrite as:

#(x^2+y^2)^(3/2) = 16x#

Subtract #16x# from both sides to get the form:

#(x^2+y^2)^(3/2) - 16x = 0#

graph{(x^2+y^2)^(3/2) - 16x=0 [-10, 10, -5, 5]}