What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {sec}^{2} θ$ $r^2+theta = -tan^2theta+4sec^2theta$ ?

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {sec}^{2} θ$ $r^2+theta = -tan^2theta+4sec^2theta$ ?

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Answer 1 · 2018-06-25T11:33:39

We know that cartesian to polar coordinates is made by

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

From these , we have then $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $θ = arctan (\frac{y}{x})$ $theta=arctan(y/x)$ and $\frac{y}{x} = tan θ$ $y/x=tantheta$ and $sec θ = \frac{r}{x} = \frac{\sqrt{x^{2} + y^{2}}}{x}$ $sectheta=r/x=sqrt(x^2+y^2)/x$

$r^{2} + θ = - {tan}^{2} θ + 4 {sec}^{2} θ$ $r^2+theta=-tan^2theta+4sec^2theta$

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

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {sec}^{2} θ$ $r^2+theta = -tan^2theta+4sec^2theta$ ?

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What is the Cartesian form of r^2+theta = -tan^2theta+4sec^2theta r2+θ=−tan2θ+4sec2θr^2+theta = -tan^2theta+4sec^2theta ?

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {sec}^{2} θ$ $r^2+theta = -tan^2theta+4sec^2theta$ ?