Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency, the discriminant $\Delta$ must be zero:
$$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$
At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
As you know, the foci of an ellipse whose equation is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$are described, if $a^2>b^2$, by the coordinates $(c,0)$ and $(-c,0)$ where $c=\sqrt{a^2-b^2}$. In fact the sum of the distances of a generical point $(x,y)$ from $(c,0)$ and $(-c,0)$ is, as we can see by using the Pythagorean theorem, $$\sqrt{(x-c)^2+(y-0)^2}+\sqrt{(x-(-c))^2+(y-0)^2}$$which we can prove to be the constant $2a$ (assuming $a>0$). In fact, if we plug $c=\sqrt{a^2-b^2}$ into the equation of the ellipse, it becomes $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$$which is equivalent to $$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$which is in turn equivalent, as we can see by adding $-2a^2cx$ to both members and rearranging the addends, to $$a^2((x-c)^2+y^2)=a^2(x^2-2cx+c^2+y^2)=a^4-2a^2cx+c^2x^2=(a^2-cx)^2$$which becomes, if we calculate the square root of both members and multiplicate by $4$ $$\pm 4a\sqrt{(x-c)^2+y^2}=4(a^2-cx)$$ but, since $c=\sqrt{a^2-b^2}<a$ and $\frac{x^2}{a^2}\le 1$ (see equation of the ellipse: $\frac{y^2}{b^2}\ge 0$) and therefore, for $x>0$, $cx\le ax\le a^2$, we can chose the sign + in front of the square root:$$0=4(a^2-cx)-4a\sqrt{(x-c)^2+y^2}$$which is in turn equivalent, as we notice if we add $x^2+2cx+c^2+y^2$ to both members, to $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2-4a\sqrt{(x-c)^2+y^2}=\left(2a-\sqrt{(x-c)^2+y^2}\right)^2$$which is finally equivalent, in turn, as we see by calculating the square root of both memebrs, to$$\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}$$where we chose the positive sign for the square root $\sqrt{(x+c)^2+y^2}$ because $2a-\sqrt{(x-c)^2+y^2}\ge 0$, in fact if $2a<\sqrt{(x-c)^2+y^2}$ then $4a^2<(x-c)^2+y^2$ and $3a^2<x^2+y^2-cx-b^2$ which is impossible because if $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $a^2>b^2$ then $\frac{x^2+y^3}{a^2}<1$. The last equality means that the sum of the distances of any point from $(c,0)$ and $(-c,0)$ is $2a$, as we wanted to show.
If $b^2>a^2$ the roles of $x$ and $y$ are inverted and therefore the foci are $(0,\sqrt{b^2-a^2})$ and $(0,-\sqrt{b^2-a^2})$.
Best Answer
There are two reasonable possibilities. One is the point where the minor axis hits the ellipse and the distance will be the semi-minor axis. The other is a point on the ellipse where the distance to a focus and the center is equal. If the distances aren't equal, you can move it toward the further and increase the minimum. It will clearly be this one if the eccentricity is high, where the distance will be essentially half the distance from the center to the focus.
Let your ellipse have semi-major axis $a$ and semi-minor axis $b$. In standard position, the equation is $(\frac xa)^2+(\frac yb)^2=1$. One point is $(0,b)$ at distance $b$. The other is at an $x$ coordinate halfway between the center and focus. A figure is below. The foci are $A,B$, a vertical dropped from $C$ bisects $BD$ and the point of interest is $C$. $f$ and $g$ have to be equal, which justifies using the bisector.
This is at $x=\frac 12\sqrt{a^2-b^2}, y=\frac ba \sqrt {a^2-\frac 14(a^2-b^2)}=\frac ba \sqrt{\frac 34a^2+\frac 14b^2}$. The distance is $\sqrt{\frac 14a^2+\frac 12b^2+\frac {b^4}{4a^2}}=\frac{a^2+b^2}{2a}$. As this is greater than $b$ it is the maximum.
Another point one might consider is the end of the ellipse where the major axis hits. Its distance from the near focus is $a - \sqrt {a^2-b^2} \lt a-\sqrt{(a-b)^2}=b$, so the other points have larger distances.