This exercise can be understood as an application of a general result about perimeter bisectors of triangles.
Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then
$$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{$\star$}$$
so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.
Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have
$$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$
Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$
The Proposition has a helpful corollary.
Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.
Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$
To solve the original problem, it basically suffices to embed the above triangle into an ellipse:
In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$
You have already made the right connection to the Director Circle. You can finish in the following way:
Let the midpoint be $(h,k)$. Then the equation of the chord using Joachimstahl notation is $$T=S_1: \dfrac{hx}{16}+\dfrac{ky}{4} = \dfrac{h^2}{16}+\dfrac{k^2}{4}$$
Let the pair of orthogonal tangents be drawn from $P(x_1,y_1)$. Then the chord of contact is $$\dfrac{xx_1}{16}+\dfrac{yy_1}{4} = 1$$
Since both represent the same straight line, we have
$$\dfrac{x_1}{h} = \dfrac{y_1}{k} = \dfrac{1}{\dfrac{h^2}{16}+\dfrac{k^2}{4}}$$
Since $P$ lies on the Director Circle $x^2+y^2 = 20$, we obtain the locus as
$$x^2+y^2 = 20 \left(\dfrac{x^2}{16}+\dfrac{y^2}{4}\right)^2$$
Best Answer
The point $(2,3)$ isn’t in the interior of the parabola, so no chord at all passes through that point. You’re correct in that regard.
On the other hand, the chord of a parabola that is bisected by a given interior point is parallel to the polar of that point. Applying this criterion the the point $(2,3)$, we can find find that its polar is $3y=2(x+2)$ and the parallel line to this through $(2,3)$ is $3y-2x=5$. This is exactly the line that you found somewhere else. Perhaps that source blindly computed this line using something like the above method without bothering to check that the point was in the interior in the first place.