How do you convert ${(x^{2} + y^{2})}^{2} = 4 (x^{2} - y^{2})$ $(x^2 + y^2)^2 = 4(x^2-y^2)$ into polar form?

Question

4255 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-31T13:48:20

$r^{2} = cos (2 θ)$ $r^2=cos(2theta)$

The pass transformations

${(x = r cos(theta)), (y = r sin(theta)):}$

give

$(r^2(sin^2(theta)+cos^2(theta)))^2=r^2cos^2(theta)-r^2sin^2(theta)$

or equivalently

$r^2=cos^2(theta)-sin^2(theta) = cos(2theta)$

See below the graphic representation.

enter image source here

How do you convert (x^2 + y^2)^2 = 4(x^2-y^2)(x2+y2)2=4(x2−y2)(x^2 + y^2)^2 = 4(x^2-y^2) into polar form?