For ((p)/(x))^2+((q)/(y))^2=(2)/(z^2) : find the complete…
For $\left(\dfrac{p}{x}\right)^2+\left(\dfrac{q}{y}\right)^2=\dfrac{2}{z^2}$: find the complete and general integrals, and the Cauchy integral through $z=x-1$ at $y=0$.
Explanation of Method
Substitute $X=x^2/2$, $Y=y^2/2$, $u=z^2/2$ to reduce the equation to the standard eikonal form $u_X^2+u_Y^2=2$, solve that by separation, then take envelopes / apply Charpit for the general and Cauchy integrals.
Step 1 · Reduce to eikonal form
With $X=x^2/2$, $Y=y^2/2$, $u=z^2/2$: $p/x=u_X$, $q/y=u_Y$ (since $dX=x\,dx$), so \[ u_X^2+u_Y^2 = 2. \]
Step 2 · Complete integral
Set $u_X=a$ ($|a|\le\sqrt2$), so $u_Y=\sqrt{2-a^2}$, and integrate: \[ u = aX+\sqrt{2-a^2}\,Y+c \implies z^2 = ax^2+\sqrt{2-a^2}\,y^2+2c. \]
Step 3 · General integral (envelope in $a$)
$\partial_a=0$ gives $x^2=ay^2/\sqrt{2-a^2}$, solving $a=\sqrt2\,x^2/\sqrt{x^4+y^4}$, $\sqrt{2-a^2}=\sqrt2\,y^2/\sqrt{x^4+y^4}$; substituting back: \[ z^2 = \sqrt{2(x^4+y^4)}+2c. \]
Step 4 · Cauchy integral through $z=x-1$ at $y=0$
In $(X,Y,u)$ coordinates the initial data is $u_0=(x_0-1)^2/2$ at $Y=0$; the strip condition gives $P_0=(x_0-1)/x_0$, and the eikonal equation gives $Q_0=\sqrt{x_0^2+2x_0-1}/x_0$. Since $P,Q$ are constant along characteristics of the eikonal equation ($\dot X=2P,\dot Y=2Q,\dot u=4$, $\dot P=\dot Q=0$), integrating and eliminating the parameter $t$ gives \[ X = X_0+\frac{P_0}{Q_0}Y,\qquad u = u_0+\frac{2Y}{Q_0}, \] which in original variables is the parametric Cauchy integral below.
Verification
Complete integral: $p=ax/z$, $q=\sqrt{2-a^2}\,y/z$,
so $(p/x)^2+(q/y)^2=(a^2+2-a^2)/z^2=2/z^2$. \checkmark
Cauchy integral: at $y=0$ the parametric formulas give $x=x_0$, $z=x_0-1=x-1$.
\checkmark; and $P_0^2+Q_0^2=[(x_0-1)^2+x_0^2+2x_0-1]/x_0^2=2x_0^2/x_0^2=2$.
\checkmark
Result
The complete, general, and Cauchy integrals are: