How do you convert #1=(-2x-3)^2+(y+1)^2# into polar form?

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Answer 1 · 2017-02-10T17:27:17

#4(r cos theta + 3/2)^2+(rsintheta+1)^2=1#.

graph{(2x+3)^2+(y+1)^2-1=0 [-3, 3, -3, 0]}

The equation has the standard form

#(x+3/2)^2/(1/2)^2+(Y+1)^2/1^2=1# that represents the ellipse, with

center #C(-3/2, -1)#, major axis along #x =-3/2#, minor axis

along #y = -1#, semi major axis a = 1,, semi minor axis #b = 1/2# and eccentricity #e= sqrt(1-b^2/a^2)=sqrt3/2.#

The conversion formula is #( x, y ) = r ( cos theta, r sin theta )#.

So, the polar equation is

#4(r cos theta + 3/2)^2+(rsintheta+1)^2=1#.