How do you convert polar equations to Cartesian equation given $r sec θ = 3$ $r sec theta = 3$ ?

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Answer 1 · 2018-06-26T11:37:37

${(x - \frac{3}{2})}^{2} + y^{2} = \frac{9}{4}$ $(x-3/2)^2 + y^2 = 9/4$

Use our polar knowledge:

$r^{2} = x^{2} + y^{2}$ $r^2 = x^2 + y^2$

$r cos θ = x and r sin θ = y$ $rcostheta = x " and " rsintheta = y$

$r sec θ = 3$ $rsectheta = 3$

$\Rightarrow \frac{r}{cos θ} = 3$ $=> r/costheta =3$

$\Rightarrow r = 3 cos θ$ $=> r = 3costheta$

$\Rightarrow r^{2} = 3 r cos θ$ $=> r^2 = 3rcostheta$

$\Rightarrow x^{2} + y^{2} = 3 x$ $=> x^2 + y^2 = 3x$

$\Rightarrow x^{2} - 3 x + \frac{9}{4} + y^{2} = \frac{9}{4}$ $=> x^2 - 3x + 9/4 + y^2 = 9/4$

$\Rightarrow {(x - \frac{3}{2})}^{2} + y^{2} = \frac{9}{4}$ $=> (x-3/2)^2 + y^2 = 9/4$

How do you convert polar equations to Cartesian equation given r sec theta = 3rsecθ=3r sec theta = 3?