Answer
Sum of the intercepts is a constant.
Work Step by Step
Our aim is to determine the tangent plane equation.
The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1)
From the given data, we have $f(x,y,z)=(x_1,y_1,z_1)$
Equation (1), becomes:
Thus,
$\dfrac{(x-x_1)}{(2\sqrt{x_1})}+\dfrac{(y-y_1)}{(2\sqrt{y_1})}+\dfrac{(z-z_1)}{(2\sqrt{z_1})}=0$
or, $\dfrac{(x-x_1)}{(\sqrt{x_1})}+\dfrac{(y-y_1)}{(\sqrt{y_1})}+\dfrac{(z-z_1)}{(\sqrt{z_1})}=0$
As we are given that the points $(x_1,y_1,z_1)$ satisfiy $\sqrt x+\sqrt y+\sqrt z=\sqrt c$
Therefore, the sum of the intercepts is a constant.
