What is the Cartesian form of $r = - {csc}^{2} θ + 4 {sec}^{2} θ$ $r = -csc^2theta+4sec^2theta$ ?

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What is the Cartesian form of $r = - {csc}^{2} θ + 4 {sec}^{2} θ$ $r = -csc^2theta+4sec^2theta$ ?

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Answer 1 · 2016-07-21T08:42:06

$x^{2} y^{2} = (4 y^{2} - x^{2}) \sqrt{x^{2} + y^{2}}$ $x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)$

Explanation:

Relation between polar coordinates $(r, θ)$ $(r,theta)$ and rectangular coordinates is given by $x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$ .

Hence, $r = - {csc}^{2} θ + 4 {sec}^{2} θ$ $r=-csc^2theta+4sec^2theta$

$\Leftrightarrow r = - \frac{1}{{sin}^{2} θ} + \frac{4}{{cos}^{2} θ}$ $hArrr=-1/sin^2theta+4/cos^2theta$ or

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

$1 = - \frac{r}{y^{2}} + \frac{4 r}{x^{2}} = \frac{r (- x^{2} + 4 y^{2})}{x^{2} y^{2}}$ $1=-r/y^2+(4r)/x^2=(r(-x^2+4y^2))/(x^2y^2)$ or

$x^{2} y^{2} = (4 y^{2} - x^{2}) \sqrt{x^{2} + y^{2}}$ $x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)$

graph{x^2y^2=(4y^2-x^2)sqrt(x^2+y^2) [-25.1, 25.08, -12.55, 12.55]}

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What is the Cartesian form of $r = - {csc}^{2} θ + 4 {sec}^{2} θ$ $r = -csc^2theta+4sec^2theta$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

What is the Cartesian form of r = -csc^2theta+4sec^2theta r=−csc2θ+4sec2θr = -csc^2theta+4sec^2theta ?

1 Answer

Explanation:

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What is the Cartesian form of $r = - {csc}^{2} θ + 4 {sec}^{2} θ$ $r = -csc^2theta+4sec^2theta$ ?