How find this $\frac{yf_{y}-z}{f_{x}}+\frac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$ solution

Question

In plane $R^3$,Find $z=f(x,y)$, such the length of the portion of any tangent line to the astroid $$z=f(x,y)$$ cut off by the coordinate axes is constant $C$,

This problem is from this post (when I answer it) How find a solution to this PDE $\frac{xf'_{x}}{f'_{y}}+\frac{yf'_{y}}{f'_{x}}+x+y=C$

my idea: I think this is anser is $$\sqrt{x}+\sqrt{y}+\sqrt{z}=C$$,because we easy to find this function is such it,

follow is my partial answer: let $z=f(x,y)$,then the tangent plane is $$f_{x}[X-x]+f_{y}[Y-y]=Z-z$$ so $$X=\dfrac{yf_{y}-z}{f_{x}}+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y,Z=z-xf_{x}-yf_{y}$$ so $$X+Y+Z=C\Longrightarrow \dfrac{yf_{y}-z}{f_{x}}+\dfrac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$$ then How find $z=f(x,y)?$ I know this PDE have one solution $$\sqrt{x}+\sqrt{y}+\sqrt{z}=C $$ is such it,But Now How prove it?

Thank you,I fell this problem is interesting. I hope someone can help me,Thank you

Answer 1

Just to bring more clarity, one another solution of $$ \frac{y f_y - z}{f_x} + \frac{x f_x -z}{f_y} - x f_x - y f_y + x+ y+ z = C $$ is $z = x + y -C$.

Answer 2

my answer is :

firstly,we assume that: $$X=\dfrac{yf_{y}-z}{f_{x}}+x=P+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y=Q+y,Z=z-xf_{x}-yf_{y}=z-R$$ then, $C^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$ $C(C-1)=(X+Y+Z)(X+Y+Z-1)=(X+Y+Z)^2-(X+Y+Z)=(x+y+z+P+Q-R)^2-(x+y+z+P+Q-R)=2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$$\Longrightarrow$$x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C^2$

$\Longrightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}=x+y+z+P+Q-R$$\Longrightarrow$$\frac{f_{x}}{f_{y}}-{f_{x}}=\frac{1}{\sqrt{x}}-1$$,\frac{f_{y}}{f_{x}}-{f_{y}}=\frac{1}{\sqrt{y}}-1$$,\frac{{1}}{f_{x}}+\frac{1}{f_{y}}=1-\frac{1}{\sqrt{z}}$$\Longrightarrow$$f_{x}=-\frac{\sqrt{z}}{\sqrt{x}},f_{y}=-\frac{\sqrt{z}}{\sqrt{y}}$

hence, $C^2=C^2$ holds