Find a point $P_2$ on an ellipse, whose chord with $P_1$ is a max distance $d$ from its nearest side I'm not sure if this solution is available in closed form, but after drawing it out I do think there will be two unique solutions always. I unfortunately have no clue where to start.
Given:

*

*An ellipse with x radius $a$, y radius $b$ and centre point $(0, 0)$

*A point on that ellipse $P_1$

*A distance $d$
Find all points $P_{2j}$ (i.e. find their coordinates) on the ellipse which satisfy the following condition (where $C$ is the chord joining $P_1$ and $P_{2j}$):

*

*The distance between $C$ and a line which is tangent to the ellipse and parallel to $C$ (on the side of the smaller segment formed by $C$ i.e. the "nearer" side) is equal to $d$
In other words, the maximum distance between $C$, and the nearer side of the ellipse from $C$, must equal $d$.
The diagram I've drawn below is an example of this. $P_{21}$ and $P_{22}$ are the desired points, whereas $P_{23}$ is an example of an invalid point. (The distances are not fully accurate)
The reason I believe the solution is closed, is because as you sweep the point $P_2j$ around the ellipse, the distance (required to be $d$) increases, reaches a turning point, and then decreases.


Here is an approach that I would take -

The equation of ellipse is $~ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
The locus of midpoints of a set of parallel chords $S$ is a straight line that passes through the center. The tangent $T$ to ellipse at the point where this diameter meets the ellipse is also parallel to the set $S$.
The given point is $P_1 (p, q)$. If $P_2(u, v)$ is the other end of the chord,
$b^2(u^2 - p^2) + a^2 (v^2 - q^2) = 0 \tag1$
Midpoint $P_m = \left(\frac{p + u}{2}, \frac{q + v}{2} \right)$
Equation of line through $P_1, P_2$,
$ (u-p) (y - q)  = (v-q) (x-p)$
$ \implies (u-p) y = (v-q)x + qu - pv$
The perpendicular distance from the center to the chord is then,
$d_m = \dfrac{|qu - pv|}{\sqrt{(u-p)^2 + (v-q)^2}}$
As $P_2$ is a point such that $P_1 P_2$ is at distance $d$ from the nearer tangent $T$ parallel to $P_1P_2$, the distance to the tangent from the center is
$d_t = \dfrac{|qu - pv|}{\sqrt{(u-p)^2 + (v-q)^2}} + d$
And if $P_t$ is the point of tangency for tangent $T$,
$ \displaystyle P_t = (x_t, y_t) = \left( \frac{d_t}{d_m} \cdot \frac{p + u}{2}, \frac{d_t}{d_m} \cdot \frac{q + v}{2} \right) ~ $ and we know,
$ \displaystyle \frac{x_t^2}{a^2} + \frac{y_t^2}{b^2} = 1 \tag2$
Using $(1)$ and $(2)$, you can obtain $P_2(u, v)$
