What the is the polar form of $y^{2} = \frac{{(x - 3)}^{2}}{y} - x^{2}$ $y^2 = (x-3)^2/y-x^2$ ?

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What the is the polar form of $y^{2} = \frac{{(x - 3)}^{2}}{y} - x^{2}$ $y^2 = (x-3)^2/y-x^2$ ?

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Answer 1 · 2018-04-22T01:02:26

$r^{3} {sin}^{3} θ - r^{2} {cos}^{2} θ - 9 + 6 r cos θ + r^{3} {cos}^{2} θ sin θ = 0$ $r^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0$

Explanation:

$y^{2} = \frac{{(x - 3)}^{2}}{y} - x^{2}$ $y^2=((x-3)^2)/y - x^2$

Put $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ ; we get :-

$r^{2} {sin}^{2} θ = \frac{{(r cos θ - 3)}^{2}}{r sin θ} - r^{2} {cos}^{2} θ$ $r^2 sin^2theta=((rcostheta-3)^2)/(rsintheta) -r^2cos^2theta$

$\Rightarrow r^{3} {sin}^{3} θ - r^{2} {cos}^{2} θ - 9 + 6 r cos θ + r^{3} {cos}^{2} θ sin θ = 0$ $rArrr^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0$

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What the is the polar form of $y^{2} = \frac{{(x - 3)}^{2}}{y} - x^{2}$ $y^2 = (x-3)^2/y-x^2$ ?

1 Answer

Explanation:

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What the is the polar form of y^2 = (x-3)^2/y-x^2 y2=(x−3)2y−x2y^2 = (x-3)^2/y-x^2 ?

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What the is the polar form of $y^{2} = \frac{{(x - 3)}^{2}}{y} - x^{2}$ $y^2 = (x-3)^2/y-x^2$ ?