Question

Question #02b85

Answer 1

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

Explanation:

One thing you can do is start by multiplying both sides of the equation `r=4/(1-cos(theta))` by `1-cos(theta)` to get `r-r cos(theta)=4`.

Next, rearrange this to get `r=4+r cos(theta)`.

Now square both sides to get `r^2=16+8r cos(theta)+r^2 cos^{2}(theta)`.

The reason this was a good idea is that you can now substitute rectangular coordinates `(x,y)` pretty quickly using the facts that `r^{2}=x^{2}+y^{2}` and `r cos(theta)=x` to get:

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

Solving this equation for `x` as a function of `y` gives

`x=(1/8)(y^2-16)=1/8 y^2-2`.

The graph of `r=4/(1-cos(theta))`, as `theta` varies over the open interval `(0,2pi)`, is the sideways parabola shown below.