How do you convert #x^2/4 + y^2 = 4# into polar form?

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How do you convert #x^2/4 + y^2 = 4# into polar form?

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Answer 1 · 2016-10-20T13:10:12

#r=4/(sqrt(cos^2theta+4sin^2theta))#

Explanation:

Let #x=rcostheta# and #y=rsintheta#
then #x^2/4+y^2=4#
#=># #(r^2cos^2theta)/4+r^2sin^2theta=4#
Simplifying
#r^2cos^2theta+4r^2sin^2theta=16#
#r^2(cos^2theta+4sin^2theta)=16#
#r^2=16/(cos^2theta+4sin^2theta)#
#r=4/(sqrt(cos^2theta+4sin^2theta))#

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How do you convert #x^2/4 + y^2 = 4# into polar form?

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