Question

How do you convert `r = 4 / (1 - cos theta)` to rectangular form?

Answer 1

`y^2-8x-16=0`

Explanation:

From the diagram we can see that point P has polar coordinates
`( r , theta )` and Cartesian coordinates `(x,y)`.

And `color(white)(88)x=rcos(theta)` , `y = rsin(theta)`

`(x,y) -> (rcos(theta), rsin(theta))`

Also:

By Pythagoras' Theorem :

`r^2=(rcostheta)^2+(rsintheta)^2`

Since:

`x=rcos(theta)` and `y = rsin(theta)`

Then:

`r^2=x^2+y^2` `:. r=sqrt(x^2+y^2)`

Using these ideas:

`r=4/(1-cos(theta))`

Substituting:

`sqrt(x^2+y^2)=4/(1-cos(theta))`

`cos(theta)=x/r`

`sqrt(x^2+y^2)=4/(1-x/r)`

`sqrt(x^2+y^2)=4/(1-x/(sqrt(x^2+y^2))`

Multiply by `(1-x/(sqrt(x^2+y^2)))`

`sqrt(x^2+y^2)-(xsqrt(x^2+y^2))/(sqrt(x^2+y^2))=(4(1-x/(sqrt(x^2+y^2))))/(1-x/(sqrt(x^2+y^2))`

`sqrt(x^2+y^2)-(xcancel(sqrt(x^2+y^2)))/(cancel(sqrt(x^2+y^2)))=(4(cancel(1-x/(sqrt(x^2+y^2)))))/((cancel(1-x/((sqrt(x^2+y^2))))))`

`sqrt(x^2+y^2)-x=4`

`sqrt(x^2+y^2)=4+x`

Squaring:

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

`y^2-8x-16=0`