I need to find tangent plane to surface $z = \frac{y^2 - 1}{x}$ that passes through $A(0,1,0)$ and $B(1,3,4)$

Normal vector of the plane I am looking for:

$$\vec{n} = \Big[\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \Big]$$

$$\vec{n} = \Big[\frac{1 - y^2}{x^2}, \frac{2y}{x}, -1 \Big]$$

Plane equation:

$$\pi: Ax + By + Cz + D = 0$$

Until now I was always given only "passing through point $P(1,2,3)$" in the task, so only one point, so I could easily find $D$ and get the final result. But here I am given two distinct points.

Not sure how to proceed.

$$\vec{n_{A}} = [0, 0 \text{#note, division by 0 here, legal?#}, -1] \quad \Rightarrow \pi_{A}: -1z+D = 0 $$

$$D = 0$$

$$\pi_{A}: -z = 0$$

And now tangent plane at point B.

$$\vec{n_{B}} = [-8, 6, -1]\quad \Rightarrow \pi_{B}: -8x + 6y -z = -D$$

$$-8 + 18 - 4 = -D$$

$$D = -6$$

$$\pi_{B}: -8x + 6y - z - 6 = 0$$

My results would probably be correct for two separate tasks, two separate planes, two separate points. But here I need one specific plane, such that it passes through both points.

Not sure what values should I plug into the normal vector $\vec{n}$. I understand that if plane passes through $A$ and $B$ then normal vector $\vec{n}$ is perpendicular to $\vec{AB}$. Also line that contains points $A$ and $B$ is contained within that plane.

Just not sure how to plug in that knowledge here. Struggling mostly with finding the normal vector. And then finding $D$.

  • $\begingroup$ Why are you computing $\vec n$ at the two points? They don’t even lie on the surface. ($A$ would, if there weren’t a slice missing from this hyperboloid along the plane $x=0$.) $\endgroup$ – amd Jun 4 at 19:51

Actually, the method that you’re trying to apply would not produce the correct answers for single point. The method finds the tangent plane to a surface at a point on the surface, but neither of the given points lies on the surface (well, $A$ would be on the hyperboloid if there weren’t a slice missing along the plane $x=0$). It makes no sense to compute $\vec n$ at these points. Instead, what you need to do is construct the equation of the tangent plane at the generic point $P_0=(x_0,y_0,z_0)$ on the surface and then plug in the coordinates of the two known points to solve for $P_0$. You’ve already worked out that $$\vec n_{P_0} = \left[{1-y_0^2\over x_0^2},2{y_0\over x_0},-1\right]$$ so the equation of the unknown plane is $\vec n_{P_0}\cdot(P-P_0)=0$. Substitute the coordinates of $A$ and $B$ for $P$ and solve the resulting system of equations for $P_0$.

Since you’ve tagged this question with “linear algebra,” here’s a linear-algebraic way to solve this. Rewrite the equation of the surface as $xz=y^2-1$ and keep in mind that the domain excludes $x=0$. This is the equation of a quadric surface which can be written in matrix form as $$\mathbf x^TQ\mathbf x = \begin{bmatrix}x&y&z&1\end{bmatrix} \begin{bmatrix} 0&0&-\frac12&0 \\ 0&1&0&0 \\ -\frac12&0&0&0 \\ 0&0&0&-1 \end{bmatrix} \begin{bmatrix}x\\y\\z\\1\end{bmatrix} = 0.$$ The equation $ax+by+cz+d=0$ of a plane can also be written as $(a,b,c,d)^T(x,y,z,1)=0$, so this plane can be represented by the vector $\mathbf\pi = (a,b,c,d)^T$. It turns out that all of the planes tangent to a nondegenerate quadric $Q$ satisfy the dual equation $\mathbf\pi^TQ^{-1}\mathbf\pi=0$ and that if $\mathbf\pi$ is tangent to $Q$, then its point of tangency is $\mathbf p = Q^{-1}\mathbf\pi$. (These equations can be derived from pole-polar relationships.)

Now, by substituting the coordinates of $A$ and $B$ into the generic plane equation, we see that the coefficients in the equations of any plane that passes through these points satisfies the system $$b+d=0 \\ a+3b+4c+d=0,$$ i.e., the representative vectors of this family of planes are the null space of $$\begin{bmatrix}0&1&0&1\\1&3&4&1\end{bmatrix}.$$ Since multiplying both sides of an equation by a nonzero constant doesn’t change the solution set, vectors that are multiples of each other represent the same plane, so to describe the family of planes through $A$ and $B$ it suffices to consider only convex combinations of basis vectors for this null space. Computing a basis for the null space via row-reduction produces $\mathbf\pi(\lambda) = (2-6\lambda,\lambda-1,\lambda,1-\lambda)^T$. So, finding the tangent planes through $A$ and $B$ becomes a matter of solving $\mathbf\pi(\lambda)^TQ^{-1}\mathbf\pi(\lambda)=0$ for $\lambda$. For this particular problem, inverting $Q$ is quite simple. This equation is quadratic in $\lambda$, so there might be two solutions. If you check where they are tangent to the surface, however, you’ll find that one of the points of tangency has the forbidden $x=0$, so you must reject that potential solution.


The general idea here is to find the equation of the tangent plane for any arbitrary point $P = (x_1, y_1, z_1)$. From here, you need to substitute both points into the equation, and solve simultaneously for $x_1, y_1$ and $z_1$.

  • $\begingroup$ Is this the equation of the tangent plane for any arbitrary point? $\frac{1 - y^2}{x^2}x + \frac{2y}{x}y - z + D = 0$ $\endgroup$ – weno Jun 4 at 11:06
  • $\begingroup$ @weno The equation of the tangent plane at $P(x_1,y_1,z_1)$ is given by$$(x-x_1,y-y_1,z-z_1)\cdot\left(\frac{1 - y_1^2}{x_1^2}, \frac{2y_1}{x_1}, -1 \right)=0$$where $(x',y',z')$ needs to be found. $\endgroup$ – Shubham Johri Jun 4 at 13:37
  • $\begingroup$ @weno Not at all. It’s not even linear. Be careful to distinguish the unknown point on the surface from the variables in the equation of the plane. $\endgroup$ – amd Jun 4 at 19:57

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