Question

Please, help toconvert the Polar Equation to Cartesian Coordinates:
\(\displaystyle{r}^{{{2}}}={\sec{{4}}}\theta\)

Answer 1

\(\displaystyle{r}^{{{2}}}={\sec{{4}}}\theta={\frac{{{1}}}{{{\cos{{4}}}\theta}}}={\frac{{{1}}}{{{\cos{{\left({2}\theta+{2}\theta\right)}}}}}}={\frac{{{1}}}{{{\cos{{2}}}\theta{\cos{{2}}}\theta-{\sin{{2}}}\theta{\sin{{2}}}\theta}}}={\frac{{{1}}}{{{1}-{8}{{\sin}^{{{2}}}\theta}+{8}{{\sin}^{{{4}}}\theta}}}}={\frac{{{1}}}{{{1}-{\frac{{{8}{y}^{{{2}}}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}+{8}{\left({\frac{{{y}^{{{2}}}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}\right)}^{{{2}}}}}}\)
\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={\frac{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}^{{{2}}}}}{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}^{{{2}}}-{8}{y}^{{{2}}}{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}+{8}{y}^{{{4}}}}}}\)
So,
\(\displaystyle{x}^{{{4}}}+{y}^{{{4}}}-{6}{x}^{{{2}}}{y}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\)

Answer 2

Thank you for the answer!