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 $y_0\approx 1.5$

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.
