finding the complete integral of a non linear pde of the first order Given is a first order non linear pde $$f\equiv16p^2z^2+9q^2+4z^2-4=0\;;\text{where $p=\frac{dz}{dx}$ and $q=\frac{dz}{dy}$}$$
I tried solving it using Charpit's method. Following were the Lagrange's auxilliary equation to find out the 2nd equation $g$.
$$\frac{dp}{p(32p^2z+8z)}=\frac{dq}{q(32p^2z+8z)}=\frac{dx}{-32pz^2}=\frac{dy}{-18q}$$
From this the second equation I got was $g\equiv \frac pq=a$, substituting the value of p from here into $f$ we get the value of q and also p as follows,$$q=\pm2\sqrt{\frac{1-z^2}{9+16a^2z^2}}\;\text{and}\;p=\pm2a\sqrt{\frac{1-z^2}{9+16a^2z^2}}\;$$for further calculations I took only the +ve signs, writing $dz=pdx+qdy$ and substituting the values the final expression turns out to be like this, $$\sqrt{\frac{9+16a^2z^2}{1-z^2}}dz=2adx+2dy$$
I am stuck here being unable to integrate the RHS of the equation. 
 A: The first step consists in replacing x bt t and y by $\lambda t$, with
$\lambda$ an arbitrary constant. By doing that we obtain the equation
\begin{equation}
\frac{d}{dt} z(t)= 4 \lambda\sqrt{\frac{1-z^2}{9+16 \lambda^2z^2}}
\end{equation}
This deals with the case $x \neq 0$, since the equations $x=t$ and $y = \lambda t$ does not allow the possibility of $x = 0$ and $y = \lambda t \neq 0$. The case $x = 0$ can also be dealt with by making the limit $x \rightarrow 0$ in solution of the above equation, since the solution of the
equation is continuous.
As the solution is concerned you can change the equation variables by replacing $t$ by $\alpha(\beta)$ and $z(t)$ by $\beta$. By doing so you get the equation 
\begin{equation}
\frac{1}{\alpha^{'}(\beta)} \sqrt{16 \lambda^2 \beta^2+9}=4 \lambda \sqrt{1-\beta^2}
\end{equation}
Which can be solved by using the incomplete elliptic integral of the second kind. So
\begin{equation}
\alpha(\beta)= \frac{3}{4 \lambda} \int_0 ^{\arcsin{\beta}} \sqrt{1-(\frac{16 \lambda^2}{9} \sin(x))^2} dx
\end{equation}
However this approach implies restictions on the possible values of $\lambda$.
