What the is the polar form of $y^{2} = \frac{{(x - 1)}^{2}}{y + x} - x^{2}$ $y^2 = (x-1)^2/(y+x)-x^2$ ?

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What the is the polar form of $y^{2} = \frac{{(x - 1)}^{2}}{y + x} - x^{2}$ $y^2 = (x-1)^2/(y+x)-x^2$ ?

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Answer 1 · 2018-04-09T01:42:54

Let $x = r cos (θ)$ $x = rcos(theta)$ and $y = r sin (θ)$ $y = rsin(theta)$

Explanation:

$r^{2} {sin}^{2} (θ) = \frac{{(r cos (θ) - 1)}^{2}}{r sin (θ) + r cos (θ)} - r^{2} {cos}^{2} (θ)$ $r^2sin^2(theta) = (rcos(theta)-1)^2/(rsin(theta)+rcos(theta)) - r^2cos^2(theta)$
$r^{2} ({sin}^{2} (θ) + {cos}^{2} (θ)) = \frac{{(r cos (θ) - 1)}^{2}}{r sin (θ) + r cos (θ)}$ $r^2(sin^2(theta)+cos^2(theta)) = (rcos(theta)-1)^2/(rsin(theta)+rcos(theta)$
$r^{3} = \frac{{(r cos (θ) - 1)}^{2}}{sin (θ) + cos (θ)}$ $r^3 = (rcos(theta)-1)^2/(sin(theta)+cos(theta))$

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What the is the polar form of $y^{2} = \frac{{(x - 1)}^{2}}{y + x} - x^{2}$ $y^2 = (x-1)^2/(y+x)-x^2$ ?

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What the is the polar form of y^2 = (x-1)^2/(y+x)-x^2 y2=(x−1)2y+x−x2y^2 = (x-1)^2/(y+x)-x^2 ?

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What the is the polar form of $y^{2} = \frac{{(x - 1)}^{2}}{y + x} - x^{2}$ $y^2 = (x-1)^2/(y+x)-x^2$ ?