I decided to tackle a problem that I thought would be a little challenging. I'm asked to show that all the tangent planes to the surface $$z = yf(\frac{x}{y})$$ pass through the origin with $f$ being a differentiable one variable funtion.

So first I moved around the surface a little to get $-z + yf\left(\frac{x}{y}\right) = 0$ and as usual, took the gradient and got

$$\left[f'\left(\frac{x}{y}\right), f\left(\frac{x}{y}\right)+\frac{x}{y}f'\left(\frac{x}{y}\right), -1\right]$$

Here is the part where I'm stuck, the tangent plane at $P = (x_0, y_0, z_0)$ will be

$$f'\left(\frac{x}{y}\right)(x-x_0) + \left[(f\left(\frac{x}{y}\right)+\frac{x}{y}f'\left(\frac{x}{y}\right)\right](y-y_0) - (z-z_0) = 0$$

But how do show in a generic way that they will ALL pass through $(0,0,0)$ without having to pick a specific point?

  • 1
    $\begingroup$ $$f'\left(\frac{x_0}{y_0}\right)(x-x_0) + \left[f\left(\frac{x_0}{y_0}\right)-\frac{x_0}{y_0}f'\left(\frac{x_0}{y_0}\right)\right](y-y_0) - (z-z_0) = 0$$ $\endgroup$
    – WW1
    Jan 31, 2017 at 4:52

1 Answer 1


Putting $x=y=z=0$ into the equation of the tangent plane, we have

\begin{array} $f'\left(\frac{x_0}{y_0}\right)(x-x_0) + \left[f\left(\frac{x_0}{y_0}\right)-\frac{x_0}{y_0}f'\left(\frac{x_0}{y_0}\right)\right](y-y_0) - (z-z_0) &=& f'\left(\frac{x_0}{y_0}\right)(-x_0) + \left[f\left(\frac{x_0}{y_0}\right)-\frac{x_0}{y_0}f'\left(\frac{x_0}{y_0}\right)\right](-y_0) - (-z_0)\\ &=&-x_0f'\left(\frac{x_0}{y_0}\right) -y_0f\left(\frac{x_0}{y_0}\right)+x_0f'\left(\frac{x_0}{y_0}\right) +z_0\\ &=& z_0-y_0f\left(\frac{x_0}{y_0}\right) \end{array} But from the given surface $z_0=y_0f\big(\frac{x_0}{y_0}\big)$ and we are done.

  • $\begingroup$ Ugh, I feel so dumb. Thank you very much $\endgroup$
    – bnoite
    Jan 31, 2017 at 21:53
  • $\begingroup$ I thought that it should be $x_0=y_0=z_0=0$. $\endgroup$
    – MathArt
    Dec 1, 2017 at 9:30

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