How do you convert #r=-4cos(theta)# to rectangular form?

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Answer 1 · 2016-02-15T07:32:12

#x^2+y^2+4x=0#

from the given, #r=-4 cos theta#

multiplying both sides of the equation by #r#

#r*r=r*(-4 cos theta)#
#r^2=-4*r cos theta#

Using #r=sqrt(x^2+y^2)# and #x=r cos theta#

it follows

#r^2=-4*r cos theta#

#(sqrt(x^2+y^2))^2=-4*x#

#x^2+y^2=-4*x#

#x^2+y^2+4x=0#

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