Hyperbolic vertices

Suppose you have two distinct points on the x-axis of the coordinate plane. If I tell you a parabola passes through those two points, where on the plane could that parabola’s vertex be? Spoiler alert: The vertex can be anywhere on the perpendicular bisector of those two points. (Neat!)

Now, suppose the two distinct points are anywhere on the coordinate plane. If I tell you that a parabola with a vertical line of symmetry passes through those two points, where on the plane could that parabola’s vertex be?

Let $P(x_1, y_1)$ and $Q(x_2,y_2)$ be the two distinct points anywhere on the coordinate plane. Let's assume that $x_1 \ne x_2 \ne 0.$ Any three points in the $xy$-plane define a unique parabola with a vertical line of symmetry that passes through all three points. So, for instance assume that the third point is $R(0,t)$ for some $t \in \mathbb{R}.$ The parabola is given by $$\begin{align*}0 &= \begin{vmatrix} x^2 & x & y & 1 \\ x_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2 & y_2 & 1 \\ 0 & 0 & t & 1 \end{vmatrix} \\ &= x^2 \left(x_1y_2 - x_2y_1 + t(x_2-x_1)\right) \\ &\quad\quad - x\left(x_1^2y_2 - x_2^2 y_1 + t(x_2^2 - x_1^2)\right) \\ &\quad\quad\quad\quad + y \left(x_1^2 x_2 - x_2^2 x_1\right) - t \left(x_1^2 x_2 - x_2^2 x_1 \right)\end{align*}$$ or equivalently $$y = \frac{x_1y_2 - x_2y_1 + t(x_2 - x_1)}{x_1x_2 (x_2 - x_1)} x^2 - \frac{x_1^2 y_2 - x_2^2 y_1 + t (x_2^2 - x_1^2)}{x_1x_2 (x_2 - x_1)} x + t.$$

The vertex of this parabola is at the point $$V(t) = V\left( \frac{x_1^2 y_2 - x_2^2 y_1 + t (x_2^2 - x_1^2) }{2 (x_1y_2 - x_2y_1 + t (x_2 - x_1))}, t - \frac{\left((x_1^2 y_2 - x_2^2 y_1 + t (x_2^2 - x_1^2) \right)^2}{4x_1x_2(x_2 - x_1) \left(x_1y_2 - x_2y_1 + t (x_2 - x_1)\right)} \right).$$ Let $x = x(t) = \frac{x_1^2 y_2 - x_2^2 y_1 + t (x_2^2 - x_1^2) }{2 (x_1y_2 - x_2y_1 + t (x_2 - x_1))},$ then we have $$y = y(t) = t - \frac{x_1y_2 - x_2y_1 + t (x_2 - x_1)}{x_1x_2(x_2-x_1)} x(t)^2.$$ By inverting $x = x(t),$ we get $$t = \frac{x_1^2 y_2 - x_2^2 y_1 - 2x(x_1y_2 - x_2y_1)}{(x_2-x_1) (2x - (x_1+x_2))},$$ so we get \begin{align*}y &= \frac{x_1^2y_2 - x_2^2y_1 - 2x(x_1y_2 - x_2y_1)}{(x_2-x_1) (2x - (x_1+x_2))} - \frac{(x_1y_2-x_2y_1) (2x - (x_1 + x_2)) - \left(x_1^2 y_2 -x_2^2 y_1 - 2x(x_1y_2 - x_2y_1)\right)}{x_1x_2(x_2 - x_1) (2x - (x_1 + x_2))} x^2 \\ &= \frac{x_1^2 y_2 - x_2^2 y_1 - 2(x_1y_2-x_2y_1) x + (y_2-y_1)x^2}{2(x_2-x_1) x - (x_2^2 - x_1^2)}\end{align*} which is a hyperbola with vertical asymptote at the midpoint between $x_1$ and $x_2$ and a second asymptote along the line $$y = \frac{1}{2} \frac{y_2-y_1}{x_2-x_1} x + \frac{(x_2y_2 - x_1y_1) -3(x_1y_2-x_2y_1)}{4(x_2-x_1)}.$$

Note that this covers the original case when $y_1 = y_2 = 0.$