Question

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

Answer 1

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

Explanation:

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]}