# Tangent plane to surface passing through two points

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 $$D$$.

• 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 $x=0$.) – 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 $$z_1$$.
• Is this the equation of the tangent plane for any arbitrary point? $\frac{1 - y^2}{x^2}x + \frac{2y}{x}y - z + D = 0$ – weno Jun 4 at 11:06
• @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. – Shubham Johri Jun 4 at 13:37