As a function of $a$, how many points are there in hyperboloid $x^2 − y^2 − z^2 = 2$ where the tangent plane is parallel to plane $z-ax=3$?

Question

PROBLEM
As a function of $a$, how many points are there in hyperboloid $x^2 − y^2 − z^2 = 2$ where the tangent plane is parallel to plane $z-ax=3$ ?

MY APPROACH
I started by finding the normal vector of the original plane $<2x, -2y, -2z>$ and the normal vector of the second plane $<-a, 0, 1>$.
I understand they have to be parallel so $<2x, -2y, -2z> \space= \space k<-a,0,1>$, but I don't know how to express it as a function of $a$.

Any help would be appreciated.

Answer 1

The plane $z-ax=\lambda$ is parallel to $z-ax=3$ for $\lambda\ne 3$. Now substituting into the hyperboloid we have

$$ x^2-y^2-(a x+\lambda)^2= 2 $$

Solving for $x$ we have

$$ x = \frac{-a\lambda\pm\sqrt{y^2(1-a^2)-2a^2+\lambda^2+2}}{a^2-1} $$

At tangency we need only a solution so

$$ y^2(1-a^2)-2a^2+\lambda^2+2=0 $$

now solving for $\lambda$ we have

$$ \lambda = \pm\sqrt{(a^2-1)(y^2+2)} $$

hence with $a = 1$ we have one $\lambda = 0$. For $|a| > 1$ we have two $\lambda$'s and for $|a| < 1$ no real $\lambda$'s

Answer 2

As you have derived

$( 2 x, -2 y, -2 z ) = k (-a, 0, 1) $

This will only occur if $(2 x , - 2 y , -2 z ) \times (-a, 0, 1) = (0,0,0) $

Expanding the left hand side

$ (-2 y , 2 a z - 2 x , - 2 a y ) = (0, 0, 0)$

Therefore, $y = 0$, and $ a z - x = 0 $ which is the line $ x = a z $ in the $xz$ plane.

Plugging this into the equation of the hyperboloid

$ a^2 z^2 - z^2 = 2 $

Hence, $ z = \pm \sqrt{\dfrac{2}{a^2 - 1} } $ , provided $a \gt 1$

Then $x = a z , y = 0 $

So there are only two points for $a \gt 1$ and no points if $ |a| \le 1 $.