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

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Answer 1 · 2016-11-25T16:33:03

$4 {(x^{2} + y^{2})}^{2} = x^{2} (x^{2} + y^{2}) \sqrt{x^{2} + y^{2}} .$ $4(x^2+y^2)^2=x^2(x^2+y^2)sqrt(x^2+y^2).$ A graph is inserted.

The conversion formula is $r (cos θ, sin θ) = (x, y)$ $r(cos theta, sin theta)=(x, y)$ , giving

$cos θ = \frac{x}{r} and sin θ = \frac{y}{r}$ $cos theta = x/r and sin theta = y/r$ , where $r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2+y^2)$ .

Substitution and reorganization gives the answer.

As r is a function of both the squares of sine and cosine, the graph

is symmetrical about $θ = 0 and θ = \frac{π}{2}$ $theta = 0 and theta = pi/2$ ( x-axis and y-axis).

graph{(x^2+y^2)^1.5(4sqrt(x^2+y^2)-x^2)-x^2y^2=0 [-80, 80, -40, 40]}

Answer 2 · 2016-11-25T16:33:38

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

The relation between polar coordinates $(r . θ)$ $(r.theta)$ and Cartesian coordinates $(x, y)$ $(x,y)$ 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 $r = - {sin}^{2} θ + 4 {sec}^{2} θ$ $r=-sin^2theta+4sec^2theta$

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

or $r^{3} x^{2} = - x^{2} y^{2} + 4 r^{4}$ $r^3x^2=-x^2y^2+4r^4$

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

What is the Cartesian form of r = -sin^2theta+4sec^2theta r=−sin2θ+4sec2θr = -sin^2theta+4sec^2theta ?