Prove that the hyperbolic paraboloid $z = xy$ and its tangent plane at an arbitrary point intersect in two perpendicular lines. (Source: Based on Shifrin's Multivariable Mathematics.)
My solution is below. I request verification and critique. In particular: Is the solution written well? How could the exposition be improved? I tried to make the solution clear and direct while still capturing the process I used to get there.
Let $(a,b,c)$ be an arbitrary point on the surface. At that point, the surface has Jacobian $\begin{bmatrix}b & a & -1\end{bmatrix}$, so the tangent plane is $$b(x-a) + a(y-b) - (z-c) = 0.$$ The intersection of the surface and its tangent is then described by this system of equations:
$$\begin{align} xy &= z \tag{1}\\ bx + ay -z - 2ab + c &= 0 \tag{2}\\ ab &= c \tag{3} \end{align}$$
Equation $(1)$ describes the surface and $(2)$ the tangent plane; $(x,y,z)$ is on their intersection iff it is a solution to both. But the tangent plane must be tangent to a point on the surface, which is a constraint provided by $(3)$.
Substituting, we reduce these to a single equation $$xy = bx + ay - ab \tag{4}.$$ Recalling that $(x-a)(y-b) = xy - bx - ay + ab$, we reduce $(4)$ to simply $$(x-a)(y-b) = 0$$ giving solutions $$[x = a, z = ay] \lor [y =b, z = bx].$$
Thus, the intersection is two perpendicular lines (parameterized by $t$): $$(a, t, at) \\(t, b, bt).$$