How do you convert $x^2-y^2=1$ to polar form?

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Answer 1 · 2016-04-28T11:36:08

$r^2=sec 2theta$ .

$(x, y) = (r cos theta, r sin theta)$

So, $x^2-y^2=r^2(cos^2theta-sin^2theta)=r^2 cos 2theta=1$ .

The semi-asymptotes are given by opposites $theta=pi/4 and theta=(5pi)/4$ and, likewise, opposites $theta=(3pi)/4 and theta=(7pi)/4$ .

In cartesian form, these equations are bundled to the simple

form $x+-y=0$ , but the rotation effect given by $theta$ is missing.. , .

How do you convert x^2-y^2=1x^2-y^2=1 to polar form?