# finding tangent planes

Question: Determine the tangent planes of $x^2+y^2-z^2=1$ at the points $(x,y,0)$ and show that they are all parallel to the $z$ axis.

Proof: Let the tangent plane of $f(x,y,z)=0$ at point $(x_{0}, y_{0},z_{0})$ be given by

$$f_x(x_{0}, y_{0},z_{0})(x-x_0)+f_y(x_{0}, y_{0},z_{0})(y-y_0)+f_z(x_{0}, y_{0},z_{0})(z-z_0)=0.$$

Now we define our function as $f(x,y,z)= x^2+y^2-z^2-1$ and $(x_{0}, y_{0},z_{0})=(x_{0}, y_{0},0)$. Finding out derivatives of $f$ we get,

$$f_x(x,y,z)=2x, \quad f_y(x,y,z)=2y, \quad f_z(x,y,z)=-2z$$

Now our derivatives in terms of $(x_{0},y_{0},0)$ are,

$$f_x(x_{0}, y_{0},0)= 2x_0, \quad f_y(x_{0}, y_{0},0)= 2y_0, \quad f_z(x_{0}, y_{0},0)=0.$$

Therefore the tangent plane is $2x_0(x-x_0)+2y_0(y-y_0)=0$, or in other words $y= \dfrac{y_0}{x_0}(x_0-x)+y_0$.

So, from this equation we see that this tangent plane is parallel to the $z$ axis.

Is this correct or am I missing information?

• Any tangent line to the unit circle will be a plane tangent to the Hyperboloid of one sheet in 3-space.
– Alan
Apr 8 '14 at 19:08
• so your saying I didn't need to show this work? Apr 8 '14 at 19:15
• Oh, no, I was looking at the surface. I think you're fine here.
– Alan
Apr 8 '14 at 19:25
• The vector of my tangent plane would be $(1,1,0)$ right showing that it is parallel to $(0,0,1)$? Apr 8 '14 at 21:38
• yes, dot product is zero.
– Alan
Apr 8 '14 at 21:39

An alternative solution, which boils down to what you've done, talking less: the gradient of $f(x,y,z) = x^2+y^2-z^2$ is normal to all the level sets of $f$. Note that: $$\langle \nabla f(x,y,0), (0,0,1)\rangle = \langle (2x,2y,0),(0,0,1)\rangle = 0,$$so if the gradient at the points $(x,y,0)$ is orthogonal to the vertical axis, then the tangent plane at these points is parallel to the axis.
This works for all of the hyperboloids $x^2+y^2-z^2 = r^2$, $r > 0$ (for $r = 0$ it degenerates to a cone, there is no tangent plane at the origin).