Tangent planes to $2+x^2+y^2$ and that contains the $x$ axis 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 $y_0\approx 1.5$
 A: 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
$$
\frac{y_0^2 + 2}{y_0} = 2y_0
\implies y_0^2 + 2 = 2y_0^2
\implies y_0^2 = 2
\implies y_0 = \pm\sqrt{2}
$$
A: 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.
