Respuesta :
Answer:
If [tex]P_0 (x_0,y_0,z_0)[/tex] is a point on the surface, then the cartesian equation of the tangent plane at [tex]P_0 (x_0,y_0,z_0)[/tex] is
[tex](\ast)z = z_0 + \frac{\partial z}{ \partial x}(x_0,y_0)\cdot (x -x_0) + \frac{\partial z}{\partial y} (x_0, y_0) (y -y_0)[/tex],
where [tex]z_0 = x_0 f \left ( \frac{y_0}{x_0}\right )[/tex].
Given that
[tex]\frac{\partial z}{\partial x} (x_0 , y_0) = f \left( \frac{y_0}{x_0}\right ) - \frac{y_0}{x_0} \cdot \frac{\partial f}{\partial x}(x_0,y_0) \ , \ \frac{\partial z}{\partial y} (x_0 , y_0)=\frac{\partial f}{\partial y} (x_0,y_0)[/tex], then
[tex](\ast)[/tex] becomes
[tex](\ast \ast) z=x_0 f \left ( \frac{y_0}{x_0}\right ) + f \left( \frac{y_0}{x_0}\right ) - \frac{y_0}{x_0}\cdot \frac{\partial f}{\partial x} (x_0,y_0)\cdot (x -x_0)+\frac{\partial f}{\partial y} (x_0,y_0)\cdot (y -y_0)[/tex].
Finally, replacing [tex] (x,y,z)=(0,0,0)[/tex] in [tex](\ast \ast)[/tex] you have that the equality is true for all [tex]P_0[/tex]. This means that [tex]O(0,0,0)[/tex]
belongs to all tangent planes and therefore, the result follows.
