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

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

How do you convert 1=(-2x-3)^2+(y+1)^21=(-2x-3)^2+(y+1)^2 into polar form?