Zop Stretching My Hyperbola!

A hyperbola in standard form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ has its focus points at $F_{1,2} = (0,\:\pm\sqrt{a^2+b^2})$ . We find the standard form of the given hyperbola $H$ with $\sec\theta=\frac{1}{\cos\theta}$ : $\begin{aligned} H: && 1&=y^2+x^2 - x^2\sec^2\theta = y^2 - x^2\tan^2\theta = \frac{y^2}{1^2} - \frac{x^2}{\ctg^2\theta}&&&\Rightarrow &&&& F_{1,2}:\: \left( 0;\:\pm\sqrt{1+\ctg^2\theta} \right)=\left( 0;\:\pm\frac{1}{|\sin\theta|} \right) \end{aligned}$ In the given domain $\theta\in \left[\frac{\pi}{8};\:\frac{\pi}{2}\right]=:D$ , sine is non-negative so we may omit the absolute values. The zop looks backwards from its position on one of the focus points towards the branches of the hyperbola. Considering it is rotated by $\theta$ counter-clockwise, the point $\vec{x}$ it sees has to satisfy the equation $\begin{aligned} \vec{x}&=\vec{F}_{1,2}-r\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}\overset{!}{\in}H,& r&>0 &&&\Rightarrow &&&& 1&\overset{!}{=}\left( \frac{\pm 1}{\sin\theta} - r\sin\theta \right)^2-(-r\cos\theta)^2\tan^2\theta = \frac{1}{\sin^2\theta}\mp 2r \end{aligned}$ Only the first solution leads to $r>0$ , so the zop must sit on the upper focus. The distance between the zop and the point $\vec{x}$ it sees is $r = \frac{1}{2}\left(\frac{1}{\sin^2\theta} -1\right)=:f(\theta)$ As we take $N$ equidistant samples with step size $\Delta\theta=\frac{3\pi}{8N}$ of this distance $f(\theta)$ and take the mean $m$ , we actually compute Riemann Sums of $f(\theta)$ : $m=\frac{1}{N}\sum_{k=0}^{N-1}f\left( \frac{\pi}{8} + \frac{3\pi}{8}\cdot\frac{k}{N} \right)=\frac{8}{3\pi}\sum_{k=0}^{N-1}f\left( \frac{\pi}{8} + \frac{3\pi}{8}\cdot\frac{k}{N} \right)\Delta\theta\quad \underset{N\rightarrow\infty}{\rightarrow}{\quad}\frac{8}{3\pi} \int_Df(\theta)\:d\theta =\frac{4}{3\pi}\left[ \ctg\theta-\theta \right]_{\frac{\pi}{8}}^{\frac{\pi}{2}}=\frac{4}{3\pi}\ctg\left( \frac{\pi}{8} \right)-\frac{1}{2}$ The answer is $p+q+r+s=4+3+8+2=\boxed{17}$