I know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.
$$\begin{pmatrix}t\\0\\1\end{pmatrix}\times
\begin{pmatrix}1\\t\\1\end{pmatrix}=
\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}$$
This could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:
\begin{align*}
a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\
ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4
\end{align*}
The $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:
$$ a^2 + ac - bc = 0 $$
Written as a matrix:
$$ (a,b,c)\cdot\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix}
\cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = 0 $$
Now, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:
$$ (x,y,1)\cdot\begin{pmatrix}1&1&0\\1&1&-2\\0&-2&0\end{pmatrix}
\cdot\begin{pmatrix}x\\y\\1\end{pmatrix} = 0 $$
Since the determinant of the upper left $2\times2$ matrix is zero, this is a parabola. It can also be described by the equation
\begin{align*}
x^2 + 2xy + y^2 &= 4y \\
(x+y)^2 &= 4y
\end{align*}
Since this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as
$$ \begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix}
\cdot\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}
=-\begin{pmatrix}2t-t^2\\t^2\\1\end{pmatrix} $$
So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.