How do you convert $r^{2} = sec 2 θ$ $r^2= sec2theta$ to polar form?

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Answer 1 · 2017-01-27T17:42:13

$x^{2} - y^{2} = 1$ $x^2-y^2=1$

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and Cartesian coordinates $(x, y)$ $(x,y)$ is

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ i.e. $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence $r^{2} = sec 2 θ$ $r^2=sec2theta$ can be written as

$r^{2} = \frac{1}{cos 2 θ} = \frac{1}{{cos}^{2} θ - {sin}^{2} θ}$ $r^2=1/(cos2theta)=1/(cos^2theta-sin^2theta)$

or $r^{2} {cos}^{2} θ - r^{2} {sin}^{2} θ = 1$ $r^2cos^2theta-r^2sin^2theta=1$

or $x^{2} - y^{2} = 1$ $x^2-y^2=1$
graph{x^2-y^2=1 [-10, 10, -5, 5]}

How do you convert r^2= sec2thetar2=sec2θr^2= sec2theta to polar form?