How do you find the rectangular equation for $r = \frac{1}{3 + sin θ}$ $r=1/(3+sintheta)$ ?

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Answer 1 · 2017-01-08T16:55:57

$y + 3 \sqrt{x^{2} + y^{2}} = 1$ $y+3sqrt(x^2+y^2)=1$

The standard polar form is

$\frac{\frac{1}{3}}{r} = 1 + \frac{1}{3} cos (θ - \frac{π}{2})$ $(1/3)/r=1+1/3cos(theta-pi/2)$

that reveals an ellipse of eccentricity $e = \frac{1}{3}$ $e = 1/3$ and and major axis

inclined at $\frac{π}{2}$ $pi/2$ to the x-axis.

The semi latus rectum $\frac{1}{3} = l = a (1 - e^{2}) = \frac{8}{9} a$ $1/3 = l = a(1-e^2)=8/9a$ .

So, major axis $2 a = \frac{3}{4}$ $2a = 3/4$ .

The conversion formula is

$r (cos θ, sin θ) = (x, y)$ $r(cos theta, sin theta ) = (x, y)$ , where $r = \sqrt{x^{2} + y^{2}} \geq 0$ $r = sqrt(x^2+y^2)>=0$ .

Making substitutions and simplifying,

$y + 3 \sqrt{x^{2} + y^{2}} = 1$ $y+3sqrt(x^2+y^2)=1$

graph{y+3sqrt(x^2+y^2)-1=0 [-1.5 1.5 -.75 .5]}

Look and name:

Look at

$r = \frac{d}{c + a cos θ + b sin θ}$ $r= d / ( c +a cos theta + b sin theta)$

and tell that this graph is an ellipse, if

$e = \frac{\sqrt{a^{2} + b^{2}}}{| c |} < 1$ $e = sqrt( a^2 + b^2 ) / (abs c) < 1$ .

For example, d = 4, c = 3, a = 2 and b = -1 gives

$r = \frac{4}{3 + 2 cos θ - sin θ}$ $r = 4 / ( 3 + 2 cos theta - sin theta)$ .

Here, $e = \frac{\sqrt{5}}{3} = 0.745 . . < 1$ $e = sqrt (5 )/ 3 = 0.745.. < 1$ , and so, it is an ellipse.

See the graph.

graph{3(x^2+y^2)^0.5+2x-y-4 = 0[-6 4 -2 4]}

How do you find the rectangular equation for r=1/(3+sintheta)r=13+sinθr=1/(3+sintheta)?