I need to find the tangent planes to $f(x,y) = 2+x^2+y^2$ and that contains the $x$ axis, so that's what I did:

$$z = z_0 + \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0)+\frac{\partial f(x_0,y_0)}{\partial y}(y-y_0) \implies \\ z = 2 + x_0^2 + y_0^2 + 2x_0(x-x_0) + 2y_0(y-y_0) \implies \\ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0$$

So since the plane must contain the $x$ axis, its normal vector must have the form $(0,y,z)$. The normal vector fot the plane I found is:

$$(2x_0, 2y_0, -1)$$

so $x_0 = 0, y_0 = y_0$

our plane has the form:

$$2y_0y -z -y_0^2+2 = 0$$

but when I plot this graph for some values of $y_0$ I only get 1 tangent plane at $y_0\approx 1.5$

  • $\begingroup$ $+2$? There must be $2yy_0 - z - y_0^2 = 0$ $\endgroup$ Aug 18 '15 at 17:14
  • $\begingroup$ @MichaelGaluza but what about $z_0$? It contains $2$ $\endgroup$ Aug 18 '15 at 17:20

So far you have worked out that the tangent plane to the surface at $(x_0,y_0,x_0^2 + y_0^2+2)$ has equation $$ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0 \tag{1} $$

If this plane contains the $x$-axis, it contains all points $(x,y,z)$ with $y=z=0$. So the equation reduces to $$ 2x x_0 = x_0^2 + y_0^2 -2 \tag{2} $$ Again, we are looking for pairs $(x_0,y_0)$ such that (2) is true for all $x$. This is different from looking for triples $(x,x_0,y_0)$ satisfying (2). If we substitute $x=\frac{x_0}{2}$ into (2) we get $$ x_0^2 = x_0^2 + y_0^2 - 2 \implies y_0^2 = 2 \implies y_0 = \pm\sqrt{2} $$ Now equation (2) reduces to $2x x_0 = x_0^2$. If we substitute $x=0$ we get $$ x_0^2 = 0 \implies x_0=0 $$ Since the points are on the paraboloid, their $z$-coordinates satisfy $$ z_0 = x_0^2 + y_0^2 + 2 = (0)^2 + \left(\sqrt{2}\right)^2 +2 = 4 $$ So there are two points on the surface for which the tangent plane contains the $x$-axis: $(0,\sqrt{2},4)$ and $(0,-\sqrt{2},4)$. We can find the equations for the planes by substituting these points into (1). The first gives $$ 2x(0) + 2y(\sqrt{2}) - z - 0^2 - 2 + 2 = 0 \implies z = 2 \sqrt{2} y $$ The second is $z = -2\sqrt{2} y$.

Another way to think about this is to project into two dimensions. It's equivalent to asking: Which lines in the $yz$-plane are tangent to $z=y^2+2$ and pass through the origin? The line through $(y_0,y_0^2+2)$ and the origin has slope $\frac{y_0^2 + 2}{y_0}$. The line through $(y_0,y_0^2+2)$ tangent to $z=y^2+2$ has slope $2y_0$. So $$ \frac{y_0^2 + 2}{y_0} = 2y_0 \implies y_0^2 + 2 = 2y_0^2 \implies y_0^2 = 2 \implies y_0 = \pm\sqrt{2} $$

  • $\begingroup$ Sorry, why did you find 2 points? And why $4$ in their coordinates? Which should be the plane equation? I'm lost. $\endgroup$ Aug 18 '15 at 18:09
  • $\begingroup$ Why did you substitute $x = \frac{x_0}{2}$? Will it work with any substitution? Will it lead to lots of different planes? I'm supposed to find all of them. $\endgroup$ Aug 18 '15 at 18:18
  • $\begingroup$ @GuerlandoOCs: I've edited answers to some of your questions into the main answer. Equation (1) is the equation of a plane: $x$, $y$, and $z$ are the variables of the plane, while $x_0$ and $y_0$ are unknowns. The condition that the plane contains the $x$-axis reduces (1) to (2), where again, $x$ is a variable and $x_0$ and $y_0$ are unknowns. Equation (2) should be true for any $x$; we choose $x = x_0/2$ and $x=0$ to find $y_0$ and $x_0$. $\endgroup$ Aug 18 '15 at 18:30
  • $\begingroup$ Couldn't I just replace $x$ by $x = \frac{x_0^2 + y_0^2 -2}{2x_0}$ in the equation, to get the general idea? Also, you found one of infinitely many possible planes, rigth? I was thinking about representing them in a general way. Can I do that? $\endgroup$ Aug 18 '15 at 18:59
  • $\begingroup$ @GuerlandoOCs You could, but then (2) reduces to $x_0^2 + y_0^2 - 2 = x_0^2 + y_0^2 - 2$, a tautology. The point is to find $x_0$ and $y_0$. Why do you think there are infinitely many possible planes? I think there are only two. $\endgroup$ Aug 18 '15 at 19:21

Taking advantage of the fact that the surface is a quadric, here’s a way to solve this without calculus. Working in homogeneous coordinates, all of the planes that contain the $x$-axis are of the form $[0:\lambda:\mu:0]$, for $\lambda$ and $\mu$ not both zero. These planes are tangent to the given surface iff they satisfy the dual conic equation $$\begin{align} \begin{bmatrix}0&\lambda&\mu&0\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-\frac12\\0&0&-\frac12&2\end{bmatrix}^{-1}\begin{bmatrix}0 \\ \lambda \\ \mu \\ 0\end{bmatrix} &= \begin{bmatrix}0&\lambda&\mu&0\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-8&-2\\0&0&-2&0\end{bmatrix}^{-1}\begin{bmatrix}0 \\ \lambda \\ \mu \\ 0\end{bmatrix} \\ &= \lambda^2-8\mu^2 = 0\end{align}$$ (the $4\times4$ matrix on the left-hand side is obtained by writing the equation $x^2+y^2-z+2=0$ in the form $\mathbf x^TC\mathbf x=0$.) therefore $\lambda=\pm2\sqrt2\mu$. Since these are homogeneous vectors, we can choose to set $\mu=1$, producing the two equations $z=\pm2\sqrt2 y$ for the tangent planes that contain the $x$-axis.


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