Question #104ce

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Question #104ce

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Answer 1 · 2017-04-04T02:31:23

$y^{2} = 3 \sqrt{x^{2} + y^{2}} + 3 x - x^{2}$ $y^2=3sqrt(x^2+y^2)+3x-x^2$

Explanation:

Use the following formulas (given by Pythagorean theorem, definition of sine, and definition of cosine, respectively):
$x^{2} + y^{2} = r^{2}$ $color(blue)(x^2+y^2=r^2)$
$y = r sin θ$ $color(blue)(y=rsintheta)$
$x = r cos θ$ $color(blue)(x=rcostheta)$

We have to manipulate the equation so that it is possible to substitute the formulas in.

(Note that the graph of $r = 3 + 3 cos θ$ $r=3+3costheta$ is a cardioid, which is not a conic section, so the rectangular form of the equation will not be in any standard form.)

$r = 3 + 3 cos θ$ $r=3+3costheta$

$r^{2} = 3 r + 3 r cos θ$ $color(blue)(r^2)=3r+3color(blue)(rcostheta)$

$x^{2} + y^{2} = 3 \sqrt{x^{2} + y^{2}} + 3 x$ $x^2+y^2=3sqrt(x^2+y^2)+3x$

$y^{2} = 3 \sqrt{x^{2} + y^{2}} + 3 x - x^{2}$ $y^2=3sqrt(x^2+y^2)+3x-x^2$

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Question #104ce

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