Two rays emanating from point $A$ are given by
$ \ell_1(t) = A + t \ v_1 $
and
$ \ell_2(s) = A + s \ v_2 $
Let $t_1, s_1$ be two given constants and let
$ B = \ell_1(t_1) $ and $ C = \ell_2(s_1) $
Let point $D$ be on $BC$ between $B$ and $C$
$ D = B + r (C - B) $ , $ 0 \lt r \lt 1 $
I want to find the (parametric) equation of the parabola that is tangent to $\ell_1, \ell_2$ and to $BC$ at $D$.
My Attempt:
The method of Bezier quadratic curve lends itself naturally to this problem. Let the tangency point with $\ell_1$ be $P_1 = \ell_1(\sigma_1) = A + \sigma_1 \ v_1 $ and let the tangency point with $\ell_2$ be $P_3 = \ell_2(\sigma_2) = A + \sigma_2 \ v_2 $, then the parametric equation of the parabola is
$ p(t) = P_1 \ (1 - t)^2 + 2 t (1 - t) \ A + P_3 \ t^2 $
At $t = t_0$, $p(t)$ passes through $D$, so
$ D = (A + \sigma_1 \ v_1 ) (1 - t_0)^2 + 2 t_0 (1 - t_0) \ A + (A + \sigma_2 \ v_2) t_0^2 $
The above equation a vector equation, so it is two scalar equations.
The third equation we can write, comes from the slope of the segment $BC$ on which $D$ lies.
The tangent vector is
$ p'(t_0) = - 2 (A + \sigma_1 v_1) (1 - t_0) + (2 - 4 t_0) \ A + 2 (A + \sigma_2 \ v_2) t_0 $
To simplify the analysis, we can shift $A$ to origin. After we find the parabola, we'll shift it back by $A$. Therefore, we can re-write the above equations with $A = 0 $, then
$ D = \sigma_1 \ v_1 (1 - t_0)^2 + \sigma_2 \ v_2 t_0^2 $
$ p'(t_0) = - 2 \sigma_1 v_1 (1 - t_0) + 2 \sigma_2 \ v_2 t_0 $
Let $w_1$ be perpendicular to $v_1$ and $w_2$ be perpendicular to $v_2$, then
$ w_1 \cdot D = \sigma_2 (w_1 \cdot v_2) \ t_0^2 $
$ w_2 \cdot D = \sigma_1 ( w_2 \cdot v_1 ) \ (1 - t_0)^2 $
Let $n$ be a normal vector to $BC$ then
$ n \cdot p'(t_0) = 0 = - 2 \sigma_1 ( v_1 \cdot n ) (1 - t_0) + 2 \sigma_2 ( v_2 \cdot n) t_0 $
Substituting for $\sigma_1$ and $\sigma_2$ from the first two equations into the third equation,
$ 0 = - \dfrac{ (w_2 \cdot D)(v_1 \cdot n) }{(w_2 \cdot v_1)( 1 - t_0 )} + \dfrac{ (w_1 \cdot D)( v_2 \cdot n )}{(w_1 \cdot v_2) t_0 } $
Muliplying through by $t_0(1 - t_0)(w_2 \cdot v_1)(w_1 \cdot v_2)$
$ - t_0 ( w_2 \cdot D)(w_1 \cdot v_2)(v_1 \cdot n) + (1 - t_0) (w_1 \cdot D) (v_2 \cdot n)(w_2 \cdot v_1) = 0 $
And this a simple linear equation in $t_0$. So now we have $t_0$, and going back to the two first equations, we can solve for $\sigma_1$ and $\sigma_2$. This fully specifies the parabola.
Any other solutions, comments, and remarks are highly appreciated.
Here's an example that uses the above method.
If $L$ and $M$ are respectively the midpoints of $P_1D$ and $P_3D$, then the midpoints $H$ and $K$ of $BL$ and $CM$ lie on the Bézier curve, at $t=t_0/2$ and $t=(1+t_0)/2$ respectively.
Hence we can write three equations: $$ \begin{align} {1\over2}B+{1\over4}D+{1\over4}P_1=& P_1\left(1-{t_0\over2}\right)^2+2At_0\left(1-{t_0\over2}\right)+P_3\left({t_0\over2}\right)^2\\ {1\over2}C+{1\over4}D+{1\over4}P_3=& P_1\left({1-t_0\over2}\right)^2+2A{1-t_0\over2}{1+t_0\over2} +P_3\left({1-t_0\over2}\right)^2\\ D= & P_1(1-t_0)^2+2At_0(1-t_0)+P_3t_0^2 \end{align} $$ which can be solved for $P_1$, $P_3$ and $D$, yielding: $$ P_1={B-At_0\over1-t_0},\quad P_3={C-A(1-t_0)\over t_0},\quad D=B(1-t_0)+Ct_0. $$ The last equality shows that $t_0=r$, thus completing the solution.