Best solutions to - "Please, help toconvert the Polar Equation to Cartesian..."

philosphy111of

Answered question

2021-11-20

Please, help toconvert the Polar Equation to Cartesian Coordinates:

r^{2} = \sec 4 θ

Answer & Explanation

Poul1963

Beginner2021-11-21Added 15 answers

r^{2} = \sec 4 θ = \frac{1}{\cos 4 θ} = \frac{1}{\cos (2 θ + 2 θ)} = \frac{1}{\cos 2 θ \cos 2 θ - \sin 2 θ \sin 2 θ} = \frac{1}{1 - 8 \sin^{2} θ + 8 \sin^{4} θ} = \frac{1}{1 - \frac{8 y^{2}}{x^{2} + y^{2}} + 8 {(\frac{y^{2}}{x^{2} + y^{2}})}^{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}

Donald Valley

Beginner2021-11-22Added 10 answers

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