Question

How do you convert `r = sec(θ + π/4)` into cartesian form?

Answer 1

`x-y=sqrt2`

Explanation:

A polar coordinate `(r,theta)` in cartesian coordinates is `(x,y)` where `x=rcostheta` and `y=rsintheta`. It is apparent that `r^2=x^2+y^2`.

Now `r=sec(theta+pi/4)=1/cos(theta+pi/4)=1/(costhetacos(pi/4)-sinthetasin(pi/4))=sqrt2/(costheta-sintheta)`

(as `sin(pi/4)=cos(pi/4)=1/sqrt2`

Hence, `rcostheta-rsintheta=sqrt2` or

`x-y=sqrt2`