<s>[INST] Show that the tangent plane to the surface $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1$ Show that the tangent plane to the surface $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1$$
At $(x_0,y_0,z_0)$ is given by $$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}=1$$
Then assume $a=b=c=1$ and show the tangent lines to the surface in the intersection with $x=x_0$ construct a cone. ($x_0 >1$).

Define $$F(x,y,z)=\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}-1$$
Then the tangent plane to the the surface at the given point is:
$$\frac{2x_{0}}{a^{2}}\left(x-x_{0}\right)-\frac{2y_{0}}{b^{2}}\left(y-y_{0}\right)-\frac{2z_{0}}{c^{2}}\left(z-z_{0}\right)=0$$
On the other hand the point is also on the surface so the equation is:
$$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}-\left(\frac{2x_{0}^{2}}{a^{2}}-\frac{2y_{0}^{2}}{b^{2}}-\frac{2z_{0}^{2}}{c^{2}}\right)=0=0$$
$$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}=1$$
Also if $a=b=c=1$ then the points on the intersection of the surface with $x=x_0$ satisfy the following relation:$$x_0^2-y^2-z^2=1$$
$$z^2+y^2=x_0^2-1\tag{1}$$
Now should I find the tangent lines to $(1)$? And show that they construct a cone?
 [/INST] Here is what the second part of the question means though it is written a bit ambiguous.
The intersection of plane $x = x_0$ and surface $x^2 - y^2 - z^2 = 1$ (as $a = b = c = 1$) is a circle. Different values of $y, z$ lead to different level curves. For example, if $y = 0$, you get a hyperbola $x^2-z^2 = 1$. Similarly if $z = 0$, you get hyperbola $x^2-y^2 = 1$. You can think of it as rotating plane $y = 0$ about x-axis. The intersection curve of a generated plane this way, with the hyperboloid, is a hyperbola. For ease of working, we parametrize the surface as,
$r(x, \theta) = (x, \sqrt{x^2-1} \cos \theta, \sqrt{x^2-1} \sin \theta), 0 \leq \theta \leq 2\pi$
For any given $\theta = \alpha$, the level curve is then given by,
$r(x) = (x, \sqrt{x^2-1} \cos \alpha, \sqrt{x^2-1} \sin \alpha)$
$r'(x) = (1, \frac{x}{\sqrt{x^2-1}} \cos \alpha, \frac{x}{\sqrt{x^2-1}} \sin \alpha)$
Equation of tangent line to the curve at $x = x_0$ is $ \ r(x_0) + r'(x_0) t$. So,
$x = x_0 + t, y = \big(\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}}\big) \cos \alpha, z = \big(\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}}\big) \sin \alpha$
Substituting, $t = x - x_0$, please note $\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}} = \frac{x_0}{\sqrt{x_0^2 - 1}} (x - \frac{1}{x_0})$
So $y^2 + z^2 = \frac{x_0^2}{x_0^2-1} (x - \frac{1}{x_0})^2$ and that is equation of a cone with its axis along x-axis and vertex at $(\frac{1}{x_0}, 0 , 0)$.
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