If a complete integral of the pde $x(p^2+q^2)=zp$, passes through the curve $x=0, z^2=4y$, then the envelope of this family passing through $x=1$ and $y=1$ has
- $z=-2$
- $z=2$
- $z=\sqrt{2+\sqrt{2}}$
- $z=-\sqrt{2+\sqrt{2}}$
As-
If the given PDE is of the form $f(x, y, z, p, q) = 0$, then Charpit's Auxiliary equations are $$\dfrac{dx}{-\frac{\partial f}{\partial p}}=\dfrac{dy}{-\frac{\partial f}{\partial q}}=\dfrac{dz}{-p \frac{\partial f}{\partial p} - q \frac{\partial f}{\partial q}}=\dfrac{dp}{\frac{\partial f}{\partial x} + p\frac{\partial f}{\partial z}}=\dfrac{dq}{\frac{\partial f}{\partial y} + q \frac{\partial f}{\partial z}}$$ For $f = p^2x+q^2x-zp$, we have $$\dfrac{dx}{-2px+z}=\dfrac{dy}{-2qx}=\dfrac{dz}{-p \frac{\partial f}{\partial p} - q \frac{\partial f}{\partial q}}=\dfrac{dp}{q^2}=\dfrac{dq}{-pq}$$ From last two equations, we get $p^2+q^2=c$ and on putting this in the main equation yields $\frac{cx}{z}=p$. Then the equation $pdx+qdy = dz$ implies $$\frac{cx}{z}dx+\sqrt{c-\frac{c^2x^2}{z^2}} dy = dz \implies zdz-cxdx= \sqrt{cz^2-c^2x^2}dy$$ Thus the solution is $$\sqrt{c}y=\sqrt{z^2-cx^2} + c_2$$ How to proceed further?