Question

How do you convert `1=(-2x-3)^2+(y+9)^2` into polar form?

Answer 1

`r^2(4cos^2theta+sin^2 theta)+6(2cos theta + 3 sin theta )+89=0`.., that can be reduced to the simple form `1/(4r)=1-sqrt3/2 cos theta`, by shifting the origin to the focus at`(-3/2, -9-sqrt3/2)`.

Explanation:

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

Any reader can easily get thecomplicated polar form

`r^2(4cos^2theta+sin^2theta)+6(2cos theta + 3 sin theta )+89=0`.

Interestingly, the given equation represents the ellipse

`(x+3/2)^2/(1/2)^2+(y+3)^2/1^2=1` that has center at (-3/2, -9), axes in e

directions of y-axis (for majoraxis) and x -axis.

Referring to a focus as pole ( origin shifted ), the polar equation

becomes as simple as `l/r = 1/(4r)=1-sqrt3/2 cos theta`