# x= r cos y = r sin `` `x^(2)` +`y^(2)` = `r^(2)` use above data to rewrite the expression in rectangular form, also need help to identify the equation as that of a line, circle,vertial parabola,...

x= r cos

y = r sin ``

`x^(2)` +`y^(2)` = `r^(2)`

use above data to rewrite the expression in rectangular form, also need help to identify the equation as that of a line, circle,vertial parabola, or horizontal parabola.

r= 4/ 1+sin `theta` (hint: sin `theta` = y/r)

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Given `r=4/(1+sintheta)` , determine the rectangular form of the function and the type of graph:

`x^2+y^2=r^2 ==>r=sqrt(x^2+y^2)`

`sin theta=y/r`

So `r=4/(1+y/r)` Multiply both sides by `1+y/r` :

`r+y=4 ==>y=4-r` Substitute for r:

`y-4=-sqrt(x^2+y^2)` Square both sides:

`y^2-8y+16=x^2+y^2`

`-8y=x^2-16`

`y=-1/8x^2+2`

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The rectangular form is `y=-1/8x^2+2` ; this is a vertical parabola opening down with vertex (0,2).

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