# Please, help toconvert the Polar Equation to Cartesian Coordinates: r2=sec4θ

philosphy111of

## Answered question

2021-11-20

Please, help toconvert the Polar Equation to Cartesian Coordinates:

${r}^{2}=\mathrm{sec}4\theta$

### Answer & Explanation

$r}^{2}=\mathrm{sec}4\theta =\frac{1}{\mathrm{cos}4\theta}=\frac{1}{\mathrm{cos}(2\theta +2\theta )}=\frac{1}{\mathrm{cos}2\theta \mathrm{cos}2\theta -\mathrm{sin}2\theta \mathrm{sin}2\theta}=\frac{1}{1-8{\mathrm{sin}}^{2}\theta +8{\mathrm{sin}}^{4}\theta}=\frac{1}{1-\frac{8{y}^{2}}{{x}^{2}+{y}^{2}}+8{\left(\frac{{y}^{2}}{{x}^{2}+{y}^{2}}\right)}^{2}$

$x}^{2}+{y}^{2}=\frac{{({x}^{2}+{y}^{2})}^{2}}{{({x}^{2}+{y}^{2})}^{2}-8{y}^{2}({x}^{2}+{y}^{2})+8{y}^{4}$

So,

$x}^{4}+{y}^{4}-6{x}^{2}{y}^{2}={x}^{2}+{y}^{2$

Thank you for the answer!

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