The apex of parabolic motion forms an ellipse of constant ellipticity.
Here's a way to generalize the geometry of the situation to non-parabola conics; unfortunately, the locus of vertices is not itself a conic (except in the parabolic case).
First, I'll reiterate OP's setup this way:
Consider the path of a projectile, subject to gravity ($g$), that leaves the origin with a speed $v$ at an angle $\theta$ with the horizontal. Parameterized by time ($t$) the path is given by $$(x,y) = \left( v t \cos\theta, -\tfrac12 gt^2 + vt \sin\theta\right) \tag{1}$$
Eliminating $t$, we find that the projectile's parabolic path has this Cartesian equation $$2 y v^2 \cos^2\theta = - g x^2 + 2 x v^2 \cos\theta \sin\theta \tag{2}$$
or, in standard form, where I'll take the opportunity to define $p := v^2/g$:
$$\left( y - \tfrac{1}{2}p\sin^2\theta \right) = - \frac{1}{2p\cos^2\theta} \left( x - \tfrac12 p \sin2\theta\right)^2 \tag{3}$$
Noting that $\sin^2\theta = \tfrac12(1-\cos2\theta)$, we see that the projectile's vertex, $(\overline{x},\overline{y})$, satisfies $$\left( \frac{2}{p} \overline{x} \right)^2 + \left(1 - \frac{4}{p} \overline{y}\right)^2 = \sin^22\theta + \cos^22\theta = 1 \quad\to\quad \frac{4}{p^2} \overline{x}^2 + \frac{16}{p^2}\left(\overline{y}-\tfrac14p\right)^2 = 1 \tag{4}$$ which describes an ellipse of eccentricity $e = \sqrt{3}/2$, independent of $p$ (and, thus, parameters $g$ and $v$). This reconfirms OP's observation.
To generalize, I'll ignore the physics and concentrate on the geometry. We can capture most of the original setup by considering a vertically-oriented conic that passes through the origin, whose tangent line there makes an angle $\theta$ with the $x$-axis. This is not enough information to determine the conic, and it also does not account for the fact that OP's parabolas get narrower as they get steeper; it's not entirely clear which additional assumption is appropriate to fill this gap. This approach looks to equation $(3)$, recalling that the reciprocal of the coefficient of the squared term is the latus rectum of the parabola; accordingly, we'll take our $\theta$-conic to have semi-latus-rectum $2\cos^2\theta$.
To apply these conditions, consider that the general equation of a vertically-oriented non-parabola conic with eccentricity $e$, major radius $a$, and center $(h,k)$ has equation $$(x-h)^2 + (1-e^2) (y-k)^2 = a^2 (1-e^2) \tag{5}$$ Then, we have the following: $$\begin{align} h^2 + (1-e^2) k^2 &= \phantom{-}a^2 ( 1 - e^2 ) & \text{(contains origin)} \tag{6a} \\ h \cos\theta &= -k (1-e^2) \sin\theta &\text{(tangent angle $\theta$ at origin)} \tag{6b} \\ \pm a(1-e^2) &= \phantom{-}p \cos^2 \theta &\text{(semi-latus-rectum length $p\cos^2\theta$)} \tag{6c} \end{align}$$ where the "$\pm$" in $(6c)$ is "$+$" for ellipses and "$-$" for hyperbolas. We can solve this system for $a$, $h$, $k$ to get $$a = \pm \frac{p\cos^2\theta}{1 - e^2} \qquad h = \pm \frac{p \sin\theta \cos^2\theta}{ \sqrt{1 - e^2 \sin^2\theta}} \qquad k = \mp \frac{p \cos^3\theta}{(1-e^2)\sqrt{1 - e^2 \sin^2\theta}} \tag{7}$$ (A second solutions switches the signs on $h$ and $k$. This corresponds to an identical conic situated on the other side of the tangent line.) Substituting into $(5)$, we have $$\left(\,x^2 + (1 - e^2) y^2\,\right) \sqrt{1 - e^2 \sin^2\theta} \;=\;\pm 2 p \cos^2\theta \left(\,x \sin\theta - y \cos\theta\,\right)\tag{8}$$ When $e=1$, taking "$\pm$" to be "$+$", this equation reduces to $(2)$, so we have successfully generalized OP's parabolas.
However, the vertices of the generalized conic have coordinates $(\overline{x}, \overline{y}) := (h, k\pm_1 a)$, where "$\pm_1$" is independent of "$\pm$". That is, $$\overline{x} = \pm \frac{p \sin\theta \cos^2\theta}{ \sqrt{1 - e^2 \sin^2\theta}} \qquad \overline{y} = \pm \frac{p \cos^2\theta \left(-\cos\theta \pm_1 \sqrt{1 - e^2 \sin^2\theta}\right)}{(1 - e^2)\sqrt{1 - e^2 \sin^2\theta}} \tag{9}$$ Eliminating the parameter $\theta$ yields
$$2 p y \left(x^2 - y^2(1-e^2)\right)^2 \pm\left(x^2 + y^2 (1-e^2)\right) \left(x^2 + y^2 (1-e)^2\right) \left(x^2 + y^2 (1+e)^2\right) = 0 \tag{10}$$
which, in general, does not itself describe a conic. (For $e=0$, this reduces to ellipse equation $(4)$, with an extra factor of $x^4$.)
Note that, since the parabola's eccentricity is $1$, it could be hiding just-about anywhere in a formula, waiting to be generalized.
As a specific example, motivated by a quantity appearing often in the above analysis, we could consider the "third" condition to be that the semi-latus rectum is not $p \cos^2\theta$, but rather $p (1 - e^2\sin^2\theta)$. The resulting vertex locus here is also generally not a conic:
$$\left(\;x^2 - (\pm 1)(\pm_1 1)\,2 p y + 4 y^2\;\right) \left(\;x^2 + y^2(1- e^2)\right) = y^2 (1 - e^2) \left(3 x^2 + 3 y^2 + e^2 y^2\right) \tag{11}$$
It's possible ---even likely--- that some other generalization gives a more-satisfying result. Indeed, one could work backwards, replacing the latus-rectum condition with an assumption that the vertex locus is some conic (or whatever curve might be desired). Such exploration is left to the reader.
Here's a better generalization than in my previous answer, using a different interpretation of that answer's equation $(3)$:
$$\left(\;y - \tfrac12\sin^2\theta\;\right) = -\frac{1}{2p\cos^2\theta}\left(\;x-\tfrac12\sin 2\theta\;\right)^2 \tag{1}$$
This time around, instead of interpreting $p\cos^2\theta$ as the semi-latus-rectum, we recognize it as twice the vertex-to-focus distance. This places the focus at $$(\hat{x},\hat{y}) = \left(\;\tfrac12 p\sin 2\theta, \tfrac12 p\sin^2\theta - \tfrac12p\cos^2\theta\;\right) = \left(\;\tfrac12 p\sin 2\theta,-\tfrac12p\cos 2\theta\;\right) \tag{2}$$ which is on a circle of radius $r := \tfrac12p$. Let's use that as an alternative "third condition" from before; since the distance from center to focus is $ae$, we can replace the earlier equation $(6a)$ with
$$h^2 + ( k + ae )^2 = r^2 \tag{3}$$
Solving the new system for $a$, $h$, $k$ gives (upon discarding a couple of extraneous cases) $$ a = \frac{\sigma r m (m + e \cos\theta)}{1 - e^2} \qquad h = \sigma r \sin\theta (m + e \cos\theta) \qquad k = - \frac{\sigma r \cos\theta (m + e \cos\theta)}{1 - e^2} \tag{4}$$ where $m := \sqrt{1-e^2\sin^2\theta}$ and $\sigma:=\pm 1$, so that the generalized conic has this equation: $$x^2 + y^2 (1- e^2) + 2 \sigma r ( x \sin\theta - y \cos\theta ) \left( m + e \cos\theta \right) = 0 \tag{5}$$ The foci have coordinates $$(\hat{x}, \hat{y}) = (h, k + \sigma_1 a) = \left(\; \sigma r\sin\theta(m+e\cos\theta), \;\frac{\sigma r (m + e \cos\theta) (\cos\theta -\sigma_1 m)}{1 - e^2}\;\right) \tag{6}$$ that satisfy
$$x^2 + y^2( 1 + \sigma_1 e)^2 + 2 \sigma \sigma_1 y r ( 1 + \sigma_1 e ) = 0 \tag{7}$$ That is,
$$\frac{x^2}{r^2} + \frac{(1 + \sigma_1 e)^2}{r^2} \left(y + \frac{\sigma\sigma_1 r}{1 + \sigma_1 e}\right)^2 = 1 \tag{8}$$
which indicates that each vertex of our conic has an elliptical locus. For $\sigma=1$, the horizontal axis is major; for $\sigma = -1$, that axis is minor. The corresponding eccentricities are
$$\sigma_1 = 1 \;:\;\frac{\sqrt{e(2+e)}}{1+e} \qquad \sigma_1 = -1\;:\; \sqrt{e(2-e)} \tag{9}$$
that, as with OP's original scenario, are independent of $r$ (or $p$, or $v$ and $g$). For $e=1$, these are $\sqrt{3}/4$ and $1$; the first confirms OP's observation, while the second is extraneous here.) $\square$
Here are some animated illustrations, showing scenarios for families of conics with eccentricity $0$, $0.5$, $0.999$, and $1.5$. (In the last case, observe that no drawing occurs while $1-e^2\sin^2\theta$ is negative.)
