Help me solve this Charpit method equation $p(1+q^2)+(b-z)q=0$

Question

The answer is given $2\sqrt{a(z-b)-1}=x+ay+c$.

But my answer is $2\sqrt{a(z-b-a)}=ax+y+c$.

Answer 1

Here $f(x,y,z,p,q)\equiv p(1+q^2)+(b-z)q$

Charpit's Method: Auxiliary equation is

$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{pf_p+qf_q}=-\frac{dp}{f_x + pf_z}=-\frac{dq}{f_y + qf_z}$

$\implies\frac{dx}{1+q^2}=\frac{dy}{2pq+b-z}=\frac{dz}{p+3pq^2+bq-zq}=-\frac{dp}{-pq}=-\frac{dq}{-q^2}$

From the last two, $-\frac{dp}{-pq}=-\frac{dq}{-q^2}$

$\implies \frac{dp}{p}=\frac{dq}{q}$

$\implies p=aq$ or, $q=ap$

If you take $p=aq$ and proceed, you will get the answer of the form $ 2\sqrt{a(z-b-a)}=ax+y+c$

Also if you take $q=ap$ and proceed, you will get the answer of the form $2\sqrt{a(z-b)-1}=x+ay+c $

In both cases your answer is true.