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

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

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Answer 1 · 2018-04-15T02:27:03

See below

Explanation:

The conversion from Rectangular to Polar:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

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

$4 r sin θ + 2 = {(3 r cos θ + 4)}^{2} + {(2 r sin θ - 1)}^{2}$ $4rsintheta+2= (3rcostheta+4)^2+(2rsintheta-1)^2$

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

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

$r (9 r {cos}^{2} θ + 24 cos θ + 4 r {sin}^{2} θ - 8 sin θ) = - 15$ $r(9rcos^2theta+24costheta+4rsin^2theta-8sintheta)=-15$

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

Let's solve the second:
$9 r {cos}^{2} θ + 4 r {sin}^{2} θ = 8 sin θ - 24 cos θ - 15$ $9rcos^2theta+4rsin^2theta=8sintheta-24costheta-15$

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

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

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

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Explanation:

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

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