I am asked to find the tangent planes to $f(x,y) = 2+x^2 + y^2$ that contain the $x$ axis. Knowing that the $x$ axis has a direction vector $\left< 1,0,0 \right>$. That said, the planes must contain that vector.
I'm not sure how I would use the plane approximation
$$L = f(x_0, y_0) + 2x (x-x_0) + 2y(y-y_0)$$
to get to the answer. The answer from the textbook is
\begin{align} ax + 2\sqrt{2} - cz + d &=0\\ ex - 2\sqrt{2} -gz + h &= 0 \end{align}
(so finding $a$, $c$, $d$, $e$, $g$ and $h$)
Any guidance is highly appreciated
The equation for the tangent plane at $(x_0,y_0,z_0)$ is:
$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\implies z=z_0+2x_0(x-x_0)+2y_0(y-y_0)$$
and it must contain $x$ axis that is $y=z=0$
$$0=z_0+2x_0(x-x_0)+2y_0(-y_0)\implies z_0-2y_0^2+2xx_0-2x_0^2=0\implies x_0=0 \quad z_0=2y_0^2$$
then
$$z=2y_0^2+2y_0(y-y_0)$$
and
$$2y_0^2=2+y_0^2\implies y_0=\pm \sqrt 2$$
that is
$$z=4+2\sqrt 2(y- \sqrt 2)$$ $$z=4-2\sqrt 2(y+ \sqrt 2)$$