 # Please, help toconvert the Polar Equation to Cartesian Coordinates: r^{2}=\sec4\theta philosphy111of 2021-11-20 Answered
Please, help toconvert the Polar Equation to Cartesian Coordinates:
${r}^{2}=\mathrm{sec}4\theta$
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${r}^{2}=\mathrm{sec}4\theta =\frac{1}{\mathrm{cos}4\theta }=\frac{1}{\mathrm{cos}\left(2\theta +2\theta \right)}=\frac{1}{\mathrm{cos}2\theta \mathrm{cos}2\theta -\mathrm{sin}2\theta \mathrm{sin}2\theta }=\frac{1}{1-8{\mathrm{sin}}^{2}\theta +8{\mathrm{sin}}^{4}\theta }=\frac{1}{1-\frac{8{y}^{2}}{{x}^{2}+{y}^{2}}+8{\left(\frac{{y}^{2}}{{x}^{2}+{y}^{2}}\right)}^{2}}$
${x}^{2}+{y}^{2}=\frac{{\left({x}^{2}+{y}^{2}\right)}^{2}}{{\left({x}^{2}+{y}^{2}\right)}^{2}-8{y}^{2}\left({x}^{2}+{y}^{2}\right)+8{y}^{4}}$
So,
${x}^{4}+{y}^{4}-6{x}^{2}{y}^{2}={x}^{2}+{y}^{2}$
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