How do you convert $r (2 - cos θ) = 2$ $r(2 - cos theta) = 2$ into a rectangular equation?

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How do you convert $r (2 - cos θ) = 2$ $r(2 - cos theta) = 2$ into a rectangular equation?

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Answer 1 · 2016-04-20T08:00:27

$3 x^{2} + 4 y^{2} - 4 x - 4 = 0$ $3 x^2+4 y^2-4 x -4 = 0$

Explanation:

$2 r = r cos θ + 2 = x + 2$ $2 r = r cos theta + 2=x+2$ .

So, $4 r^{2} = {(x + 2)}^{2}$ $4 r^2=(x+2)^2$ .
$4 (x^{2} + y^{2}) = x^{2} + 4 x + 4$ $4 (x^2+y^2)=x^2+4x+4$
$3 x^{2} + 4 y^{2} - 4 x - 4 = 0$ $3 x^2+4 y^2-4 x -4 = 0$

The given equation is $\frac{1}{r} = 1 - (\frac{1}{2}) cos θ$ $1/r=1-(1/2) cos theta$
This represents an ellipse with eccentricity $e = \frac{1}{2}$ $e = 1/2$ and semi-latus rectum l = 1. Semi-major axis a = $\frac{l}{1 - e^{2}} = \frac{4}{3}$ $l/(1-e^2)=4/3$ .
The negative sign indicates that the initial line is reversed, from pole (a focus) to the center of the ellipse.

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How do you convert $r (2 - cos θ) = 2$ $r(2 - cos theta) = 2$ into a rectangular equation?

1 Answer

Explanation:

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How do you convert r(2 - cos theta) = 2r(2−cosθ)=2 r(2 - cos theta) = 2 into a rectangular equation?

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How do you convert $r (2 - cos θ) = 2$ $r(2 - cos theta) = 2$ into a rectangular equation?