[Math] Sum of $x, y, z$ intercepts of a tangent plane is constant.

multivariable-calculus

Show that the sum of the $x$-, $y$- and $z$-intercepts of any tangent plane
to the surface $\sqrt x +\sqrt y +\sqrt z = 1$ is a constant.

I tried making an equation: $z = (1-\sqrt{x}+\sqrt{y})^2$ and I followed the standards steps of finding an tangent plane through the partial derivatives with respect to $x$ and $y$. I let $x_0, y_0, z_0$ be any point on the surface.
To find each intercept, I substitute $0$ for the other two variables. All I got are complicated equations.

Any ideas?

Best Answer

You need to write equation of plane in implicit form ie $F(x,y,z) = \sqrt x + \sqrt y + \sqrt z - 1 = 0$. Then you can write equation of tangent as $F_x(x,y,z) (x-x_0) + F_y (x,y,z)(y - y_0) + F_z (x,y,z)(z-z_0) = 0 $. And finally substitute suitable zeros (eg y =0, z=0 for x-intercept) to find all 3 intercepts. When you add them it would come as some function of $(\sqrt x + \sqrt y + \sqrt z)$, hence constant.