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 · 2016-04-21T03:56:50

$r^{2} = sec 2 θ$ $r^2=sec 2theta$ .

Explanation:

$x = r cos θ and y = r sin θ$ $x=r cos theta and y = r sin theta$ . Also, use ${cos}^{2} θ - {sin}^{2} θ = cos 2 θ$ $cos^2theta-sin^2theta=cos 2theta$ .

The given equations becomes $r^{2} cos 2 θ = 1 . S o, r^{2} = sec 2 θ$ $r^2 cos 2theta=1. So, r^2=sec 2theta$ .

This represents a rectangular hyperbola with asymptotes along lines given by $r^{2} = \infty = sec 2 θ . S o, θ = \pm \frac{π}{4} and θ = \pm \frac{3 π}{4}$ $r^2=oo=sec 2theta. So, theta = +- pi/4 and theta=+-(3pi)/4$ , give the asymptotes as four half lines, from the the pole which is thw center of the hyperbola.

For a rectangular hyperbola, the semi- major axis a = semi-transverse axis b.
The eccentricity of a rectangular hyperbola = $\sqrt{2}$ $sqrt 2$ . .

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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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Explanation:

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