Transform the following cartesian equations in polar equations

Question

$$4y^2-20x-25=0$$ The answer given by the textbook is $r=\frac{5}{2(1-\cos \theta)}$ and I couldn't get to this result.

I have done $x=r\cos\theta$ and $y=r\sin\theta$ and it leads to $4r^2\sin^2\theta-20r\cos\theta-25=0$ and then I tried different paths but none leads me to the result given.

Appreciate any help.

Answer 1

$4r^2\sin^2\theta-20r\cos\theta-25=0$

$\Rightarrow4r^2(1-\cos^2\theta)-20r\cos\theta-25=0$

$\Rightarrow4r^2\cos^2\theta+20r\cos\theta+25=4r^2$

$\Rightarrow(2r\cos\theta+5)^2=(2r)^2$

$\Rightarrow2r\cos\theta+5=2r$

$\Rightarrow r=\dfrac{5}{2(1-\cos\theta)}$