xy - 2x = 0
Convert the rectangular equation into polar form.
We first parametrize it
$\displaystyle x=r\cos(\theta)$
and $\displaystyle y=r\sin(\theta)$
Inputting this we get
$\displaystyle (r\cos(\theta))(r\sin(\theta))-2r\cos(theta)=0$
This can be rewritten as
$\displaystyle \frac{r^2}{2}\sin(2\theta)-2r\cos(\theta)=0$
Or we could have factored out a $\displaystyle r\cos(\theta)$
$\displaystyle r\cos(\theta)(r\sin(\theta)-1)=0$
Or finally we can solve this using the quadratic equation
If we let $\displaystyle u=\cos(\theta)\Rightarrow\sqrt{1-u^2}=\sin(\theta)$
so we have
$\displaystyle r^2u\sqrt{1-u^2}-2ru$
Now using the quadratic equation we get
$\displaystyle r=\frac{-u\sqrt{1-u^2}\pm\sqrt{u^2(1-u^2)-4(-2u)}}{2}$
So $\displaystyle r=\frac{-\cos(\theta)\sin(\theta)\pm\sqrt{\cos^2(\theta)\si n^2(\theta)+8\cos(\theta)}}{2}$
xy-2x=0
xy=2x
y=2, x does not equal zero
OR
xy-2x=0
t=theta
x=r*cos(t), -pi < t <pi
y=r*sin(t)
xy-2x=r*sin(t)*r*cos(t)-2*r*cos(t)
r*sin(t)*r*cos(t)-2*r*cos(t)=0
r*sin(t)*r*cos(t)=2*r*cos(t)
r*sin(t)*cos(t)=2*cos(t), r does not equal zero
r=sec(t)*csc(t)*(2*cos(t))
r=2*sec(t)*csc(t)*[cos(t)]
maybe this is where the computation in the book went wrong, you can see it's the same except for a factor of cos(t). Simplifying we get,
r=2*csc(t)
r=2/sin(t), r does not equal zero, -pi < t <pi
Even editors make mistakes