How do you convert $9=(4x+7)^2+(6y-9)^2$ into polar form?

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Answer 1 · 2016-05-19T17:31:41

$r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0$

The relation between polar coordinates $(r,theta)$ and Cartesian coordinates $(x,y)$ is given by

$x=rcostheta$ , $y=rsintheta$ , $r^2=x^2+y^2$ .

Using them $(4x+7)^2+(6y-9)^2=9$ is

$(4x+7)^2+(6y-9)^2=9$

$(4rcostheta+7)^2+(6rsintheta-9)^2=9$

or $16r^2cos^2theta+56rcostheta+36r^2sin^2theta-108rsintheta+81=9$

or $16r^2+56rcostheta+20r^2sin^2theta-108rsintheta+72=0$

or $4r^2+14rcostheta+5r^2sin^2theta-27rsintheta+18=0$

or $r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0$

How do you convert 9=(4x+7)^2+(6y-9)^29=(4x+7)^2+(6y-9)^2 into polar form?