Question

How do you convert `r = -3sec(theta)` to rectangular form?

Answer 1

`x=-3` The graph is a vertical line.

Explanation:

`sec(theta) = 1/cos(theta)`

And cosine in `("adjacent")/("hypotenuse")`

So `sec(theta) =1/cos(theta)=("hypotenuse")/("adjacent")`

But hypotenuse`= sqrt(x^2+y^2) " and adjacent "= x`

so `sec(theta)=sqrt(x^2+y^2)/x`

So we now have:`" "r=-3sqrt(x^2+y^2)/x`

But `" "r=sqrt(x^2+y^2)` giving:

`" "sqrt(x^2+y^2) =-3sqrt(x^2+y^2)/x`

Divide both sides by `sqrt(x^2+y^2)` to get

`1=-3/x`

Multiply by `x` to finish:

`x=-3`

Alternatively

Use `x=rcostheta` and `sectheta = 1/costheta`

So we have

`r=-3sectheta`

`r = (-3)/costheta`

`rcostheta = -3`

`x=-3`