How do you convert $r (2 + cos θ) = 1$ $r(2 + costheta) = 1$ into cartesian form?

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How do you convert $r (2 + cos θ) = 1$ $r(2 + costheta) = 1$ into cartesian form?

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Answer 1 · 2017-02-08T12:30:08

$\frac{{(x + \frac{1}{3})}^{2}}{\frac{3}{4}} + \frac{y^{2}}{\frac{1}{3}} = 1$ $(x+1/3)^2/(3/4)+y^2/(1/3)=1$ See the markings at the center and the pole of the graph of this ellipse.

Explanation:

graph{(2sqrt(x^2+y^2)+x-1)(x^2+y^2-.0005)((x+1/3)^2+y^2-.0005)=0 [-1.25, 1.25, -0.625, 0.625]} The equation has the form

$\frac{\frac{1}{2}}{r} = 1 + \frac{1}{2} cos θ$ $(1/2)/r=1+1/2costheta$ .

This represents the ellipse with focus S as pole, $θ = 0$ $theta=0$ for S-side

major axis, eccentricity $e = \frac{1}{2}$ $e = 1/2$ and semi major axis a = 2/3, from

$a (1 - e^{2}) = l = \frac{1}{2}$ $a(1-e^2)=l=1/2$ .

The conversion formula is

$r (cos θ, sin θ) = (x, y)$ $r(costheta, sin theta)=(x, y)$ .

So, $2 r + r cos θ = 2 \sqrt{x^{2} + y^{2}} + x = 1$ $2r+rcostheta=2sqrt(x^2+y^2)+x=1$ , giving

$x^{2} + y^{2} = \frac{1}{4} {(1 - x)}^{2} = \frac{x^{2}}{4} - \frac{x}{2} + \frac{1}{4}$ $x^2+y^2=1/4(1-x)^2=x^2/4-x/2+1/4$ .

In the standard form, this is

$\frac{{(x + \frac{1}{3})}^{2}}{\frac{3}{4}} + \frac{y^{2}}{\frac{1}{3}} = 1$ $(x+1/3)^2/(3/4)+y^2/(1/3)=1$

Answer 2 · 2017-02-08T12:57:46

$3 x^{2} + 4 y^{2} + 2 x = 1$ $3x^2+4y^2+2x=1$

Explanation:

In converting from polar $(r, θ)$ $(r,theta)$ to Cartesian $(x, y)$ $(x,y)$
we have the relations:
$XXX r = \sqrt{x^{2} + y^{2}}$ $color(white)("XXX")color(red)(r=sqrt(x^2+y^2))$
and
$XXX cos (θ) = \frac{x}{\sqrt{x^{2} + y^{2}}}$ $color(white)("XXX")color(blue)(cos(theta)=x/sqrt(x^2+y^2))$
(and others)

Polar form: $r (2 + cos (θ)) = 1$ $color(red)(r)(2+color(blue)(cos(theta)))=1$
becomes (in Cartesian form)
$XXX \sqrt{x^{2} + y^{2}} \cdot (2 + \frac{x}{\sqrt{x^{2} + y^{2}}}) = 1$ $color(white)("XXX")color(red)(sqrt(x^2+y^2)) * (2 +color(blue)(x/sqrt(x^2+y^2)))=1$

$XXX \to 2 \sqrt{x^{2} + y^{2}} + x = 1$ $color(white)("XXX")rarr 2sqrt(x^2+y^2)+x=1$

$XXX \to \sqrt{x^{2} + y^{2}} = \frac{1 - x}{2}$ $color(white)("XXX")rarr sqrt(x^2+y^2)=(1-x)/2$

$XXX \to x^{2} + y^{2} = \frac{1 - 2 x + x^{2}}{4}$ $color(white)("XXX")rarr x^2+y^2=(1-2x+x^2)/4$

$XXX \to 4 x^{2} + 4 y^{2} = 1 - 2 x + x^{2}$ $color(white)("XXX")rarr 4x^2+4y^2=1-2x+x^2$

$XXX \to 3 x^{2} + 4 y^{2} + 2 x = 1$ $color(white)("XXX")rarr 3x^2+4y^2+2x=1$

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How do you convert $r (2 + cos θ) = 1$ $r(2 + costheta) = 1$ into cartesian form?

2 Answers

Explanation:

Explanation:

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Science

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Humanities

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How do you convert r(2 + costheta) = 1r(2+cosθ)=1r(2 + costheta) = 1 into cartesian form?

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Explanation:

Explanation:

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How do you convert $r (2 + cos θ) = 1$ $r(2 + costheta) = 1$ into cartesian form?