# A polar conic section Consider the equation r^2 = sec 2 theta. Convert the equation to Cartesian coordinates and identify the curve

A polar conic section Consider the equation ${r}^{2}=sec2\theta$. Convert the equation to Cartesian coordinates and identify the curve
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Given: we have an equation in polar form ${r}^{2}=\mathrm{sec}2\theta$ Part a) to determinate the equation to Cartesian coordinates and identify the curve Explain: to convert the equation we can write the equation sa follows $r=\sqrt{\mathrm{sec}2\theta }$
$r=1\sqrt{\mathrm{cos}2\theta }$ We know these relations $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta ,r=\sqrt{{x}^{2}+{y}^{2}}$ Can be written $\mathrm{cos}\theta$ $=x/r,\mathrm{sin}\theta =y/r$ $r=\frac{1}{\sqrt{{\mathrm{cos}}^{2}\theta }}=\frac{1}{\sqrt{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }}\left[{\mathrm{cos}}^{2}\theta ={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right]$

Putting in the equation we have $r={\frac{1}{\sqrt{\frac{x}{r}}}}^{2}-{\frac{y}{r}}^{2}$ $\frac{{x}^{2}}{{r}^{2}}-\frac{{y}^{2}}{{r}^{2}}=\frac{1}{{r}^{2}}$ We have the equation ${x}^{2}-{y}^{2}=1$