How do you convert $2 x^{2} = - 2 y^{2}$ $2x^2=-2y^2$ to polar form?

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Answer 1 · 2016-05-23T14:27:52

r = 0 and theta is arbitrary.

As $(x, y) = (r cos θ, r sin θ)$ $(x, y)=(r cos theta, r sin theta)$ , the given equation reduces to

$r^{2} ({cos}^{θ} + {sin}^{2} θ) = r^{2} (1) = r^{2} = 0$ $r^2(cos^theta+sin^2 theta)=r^2(1)=r^2=0$ , for every $θ$ $theta$ .

So, r =0 and $θ$ $theta$ is arbitrary.

Interestingly, this is a proof for "0-dimensional point in space is a null

space vector#..

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