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

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

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Answer 1 · 2016-04-28T11:36:08

$r^{2} = sec 2 θ$ $r^2=sec 2theta$ .

Explanation:

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

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

The semi-asymptotes are given by opposites $θ = \frac{π}{4} and θ = \frac{5 π}{4}$ $theta=pi/4 and theta=(5pi)/4$ and, likewise, opposites $θ = \frac{3 π}{4} and θ = \frac{7 π}{4}$ $theta=(3pi)/4 and theta=(7pi)/4$ .

In cartesian form, these equations are bundled to the simple

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

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

1 Answer

Explanation:

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