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

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Answer 1 · 2017-01-08T16:55:57

#y+3sqrt(x^2+y^2)=1#

The standard polar form is

#(1/3)/r=1+1/3cos(theta-pi/2)#

that reveals an ellipse of eccentricity #e = 1/3# and and major axis

inclined at #pi/2# to the x-axis.

The semi latus rectum #1/3 = l = a(1-e^2)=8/9a#.

So, major axis #2a = 3/4#.

The conversion formula is

#r(cos theta, sin theta ) = (x, y)#, where #r = sqrt(x^2+y^2)>=0#.

Making substitutions and simplifying,

#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= d / ( c +a cos theta + b sin theta)#

and tell that this graph is an ellipse, if

#e = sqrt( a^2 + b^2 ) / (abs c) < 1#.

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

#r = 4 / ( 3 + 2 cos theta - sin theta)# .

Here, #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]}