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

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

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Answer 1 · 2016-12-22T05:51:54

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

Explanation:

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

Any reader can easily get thecomplicated polar form

$r^{2} (4 {cos}^{2} θ + {sin}^{2} θ) + 6 (2 cos θ + 3 sin θ) + 89 = 0$ $r^2(4cos^2theta+sin^2theta)+6(2cos theta + 3 sin theta )+89=0$ .

Interestingly, the given equation represents the ellipse

$\frac{{(x + \frac{3}{2})}^{2}}{{(\frac{1}{2})}^{2}} + \frac{{(y + 3)}^{2}}{1^{2}} = 1$ $(x+3/2)^2/(1/2)^2+(y+3)^2/1^2=1$ 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 $\frac{l}{r} = \frac{1}{4 r} = 1 - \frac{\sqrt{3}}{2} cos θ$ $l/r = 1/(4r)=1-sqrt3/2 cos theta$

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

1 Answer

Explanation:

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How do you convert 1=(-2x-3)^2+(y+9)^21=(−2x−3)2+(y+9)21=(-2x-3)^2+(y+9)^2 into polar form?

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