How do you convert $r = 4 sec (θ) (1 + tan (θ))$ $r=4sec(theta)(1+ tan(theta))$ into cartesian form?

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Answer 1 · 2016-10-26T09:53:48

$\frac{1}{4} = \frac{1}{x} + \frac{1}{y}$ $1/4=1/x+1/y$

Polar coordinates $(r, θ)$ $(r,theta)$ and Cartesian coordinates $(x, y)$ $(x,y)$ are related as

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ , i.e. $tan θ = \frac{y}{x}$ $tantheta=y/x$

Hence, $r = 4 sec θ (1 + tan θ)$ $r=4sectheta(1+tantheta)$ can be written as

$r cos θ = 4 (1 + tan θ)$ $rcostheta=4(1+tantheta)$

or $y = 4 (1 + \frac{y}{x})$ $y=4(1+y/x)$

or $x y = 4 x + 4 y$ $xy=4x+4y$

or $\frac{x y}{4 x y} = \frac{4 x}{4 x y} + \frac{4 y}{4 x y}$ $(xy)/(4xy)=(4x)/(4xy)+(4y)/(4xy)$

or $\frac{1}{4} = \frac{1}{y} + \frac{1}{x}$ $1/4=1/y+1/x$ or $\frac{1}{4} = \frac{1}{x} + \frac{1}{y}$ $1/4=1/x+1/y$

How do you convert r=4sec(theta)(1+ tan(theta))r=4sec(θ)(1+tan(θ))r=4sec(theta)(1+ tan(theta)) into cartesian form?