How do you convert $4y+2=(3x+4)^2+(2y-1)^2$ into polar form?

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Answer 1 · 2018-04-15T02:27:03

See below

The conversion from Rectangular to Polar:
$x=rcostheta$
$y=rsintheta$

Substitute for $x$ and $y$ and solve for $r$ :
$4(rsintheta)+2= (3(rcostheta)+4)^2+(2(rsintheta)-1)^2$

$4rsintheta+2= (3rcostheta+4)^2+(2rsintheta-1)^2$

$4rsintheta+2= 9r^2cos^2theta+24rcostheta+16+4r^2sin^2theta-4rsintheta+1$

$9r^2cos^2theta+24rcostheta+4r^2sin^2theta-8rsintheta+15=0$

$r(9rcos^2theta+24costheta+4rsin^2theta-8sintheta)=-15$

At this point either $r$ is equal to $-15$ or $(9rcos^2theta+24costheta+4rsin^2theta-8sintheta)=-15$

Let's solve the second:
$9rcos^2theta+4rsin^2theta=8sintheta-24costheta-15$

$r(9cos^2theta+4sin^2theta)=8sintheta-24costheta-15$

$r=(8sintheta-24costheta-15)/(9cos^2theta+4sin^2theta)$

How do you convert 4y+2=(3x+4)^2+(2y-1)^24y+2=(3x+4)^2+(2y-1)^2 into polar form?