method of characteristics for non-linear PDE

I'm trying to solve the PDE $u_x^2-u_y^2=8u$ with initial conditions $u(x,x)=f(x)$. I have that $F(x,y,u,p,q)=p^2-q^2-8u$, with $p=u_x, q=u_y$, and then \begin{equation*} \begin{array}{ll} x'(s)=2p\\y'(s)=-2q\\u'(s)=2p^2-2q^2=16u\\ p'(s)=8p\\q'(s)=8q \end{array} \end{equation*} so \begin{equation*} \begin{array}{ll} p(s)=c_1e^{8s}\\q(s)=c_2e^{8s}\\x(s)=\frac{1}{4}c_1e^{8s}+c_3\\y(s)=\frac{1}{4}c_2e^{8s}+c_4\\u(s)=c_5e^{16s} \end{array} \end{equation*} But I'm not sure how to go from here. In particular, how should I apply the initial condition?

I appreciate the help. Thanks!

Hint:

Let $u=v^2$ ,

Then $u_x=2vv_x$

$u_y=2vv_y$

$\therefore(2vv_x)^2-(2vv_y)^2=8v^2$ with $v(x,x)=\pm\sqrt{f(x)}$

$4v^2v_x^2-4v^2v_y^2=8v^2$ with $v(x,x)=\pm\sqrt{f(x)}$

$v_x^2-v_y^2=2$ with $v(x,x)=\pm\sqrt{f(x)}$

Let $\begin{cases}p=\dfrac{x+y}{2}\\q=\dfrac{x-y}{2}\end{cases}$ ,

Then $v_x=v_pp_x+v_qq_x=\dfrac{v_p+v_q}{2}$

$v_y=v_pp_y+v_qq_y=\dfrac{v_p-v_q}{2}$

$\therefore\dfrac{(v_p+v_q)^2}{4}-\dfrac{(v_p-v_q)^2}{4}=2$ with $v(p,0)=\pm\sqrt{f(p)}$

$\dfrac{v_p^2+2v_pv_q+v_q^2-v_p^2+2v_pv_q-v_q^2}{4}=2$ with $v(p,0)=\pm\sqrt{f(p)}$

$v_pv_q=2$ with $v(p,0)=\pm\sqrt{f(p)}$

$v_q-\dfrac{2}{v_p}=0$ with $v(p,0)=\pm\sqrt{f(p)}$

$\dfrac{2v_{pp}}{(v_p)^2}+v_{pq}=0$ with $v(p,0)=\pm\sqrt{f(p)}$

Let $w=v_p$ ,

Then $\dfrac{2w_p}{w^2}+w_q=0$ with $w(p,0)=\pm\dfrac{f_p(p)}{2\sqrt{f(p)}}$

$\dfrac{dq}{dt}=1$ , letting $q(0)=0$ , we have $q=t$
$\dfrac{dw}{dt}=0$ , letting $w(0)=w_0$ , we have $w=w_0$
$\dfrac{dp}{dt}=\dfrac{2}{w^2}=\dfrac{2}{w_0^2}$ , letting $p(0)=F(w_0)$ , we have $p=\dfrac{2t}{w_0^2}+F(w_0)=\dfrac{2q}{w^2}+F(w)$ , i.e. $w=G\left(p-\dfrac{2q}{w^2}\right)$
$w(p,0)=\pm\dfrac{f_p(p)}{2\sqrt{f(p)}}$ :
$G(p)=\pm\dfrac{f_p(p)}{2\sqrt{f(p)}}$