Show that all the tangent plans to the conic surface $z = xf(\frac{y}{x})$ at the point $M(x_o,y_o,z_o)$, where $x_o \neq 0$, pass through the origin of the cordinates
First, I've found the tangent plan at this generic point $M$ of the surface:
$z – z_o = (x-x_o)(f(\frac{x_o}{y_o}) – x_o \frac{y_o}{x_o^2}f'_x(\frac{y_o}{x_o})) + (y-y_o)(x_o \frac{1}{x_o} f'_y(\frac{y_o}{x_o}))$
We want to check if $(0,0,0)$ is a point of this plan. Then:
$-z_o = -x_o f(\frac{y_o}{x_o}) + y_o f'_x(\frac{y_o}{x_o}) – y_o f'_y(\frac{y_o}{x_o})$
But I couldn't prove that this equality holds (i.e,$ y_o f'_y(\frac{y_o}{x_o}) = y_o f'_x(\frac{y_o}{x_o})$)
Thanks in advance!
Best Answer
I sense an apparent misconception on your part regarding the term $f(y/x)$. This is hiding a second function $g: \mathbb{R}^2 \to \mathbb{R}$ defined as $g(x,y) = f(y/x)$, with $f: \mathbb{R} \to \mathbb{R}$. (Possibly not all of $\mathbb{R}^2$ or $\mathbb{R}$, in which case just assume they are defined for the largest subset avaiable for the exercise.)
Therefore, expressions as $f'_y$ and $f'_x$ don't make much sense, because $f$ is not a function of two variables, but $g$, as we defined it, is.
Calculating the partial derivatives of $z$, we find
$$\begin{align} \frac{\partial z}{\partial x} & = f \left( \frac{y}{x} \right) + x f' \left( \frac{y}{x} \right) \left( - \frac{1}{x^2} \right) = f \left( \frac{y}{x} \right) - \frac{y}{x} f' \left( \frac{y}{x} \right), \\ \frac{\partial z}{\partial y} & = x f' \left( \frac{y}{x} \right) \frac{1}{x} = f' \left( \frac{y}{x} \right). \end{align}$$
Notice also that $z_0 = x_0 f(y_0/x_0)$. Putting it all together in the tangent plane equation, we have
$$z - x_0 f \left( \frac{y_0}{x_0} \right) = \left( f \left( \frac{y_0}{x_0} \right) - \frac{y_0}{x_0} f' \left( \frac{y_0}{x_0} \right) \right) (x - x_0) + \left( f' \left( \frac{y_0}{x_0} \right) \right) (y-y_0).$$
Expand out and you'll simplify it down to
$$z = \left( f \left( \frac{y_0}{x_0} \right) - \frac{y_0}{x_0} f' \left( \frac{y_0}{x_0} \right) \right) x + \left( f' \left( \frac{y_0}{x_0} \right) \right) y.$$
We see then that $x=y=z=0$ is a solution to this plane equation and therefore it passes through the origin.