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Given an ellipse $E$ (with the foci $f_1$ and $f_2$ and the center $c$), and a point $p$, which is the maximum distance that $p$ can be to all these 3 points to be within the ellipse $E$?

I.e., which is the maximum value for $min(dist(p,f_1),dist(p,f_2),dist(p,c))$ such that $p$ is within $E$?

Thanks!

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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 $B,D$, $EF$ bisects $BD$ and the point of interest is $F$. $g$ and $h$ have to be equal, which justifies using the bisector. The coordinates below assume the ellipse is centered, so subtract $4$ from all the $x$ coordinates in the figure.

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.

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  • $\begingroup$ @achillehui: you are right. Thanks $\endgroup$ – Ross Millikan Jun 17 '16 at 1:55
  • $\begingroup$ I found $y$ by taking the $x$ value and solving for the point on the ellipse. If you have $x$, check that your point is on the ellipse, because it can only be on the border. $\endgroup$ – Ross Millikan Jun 17 '16 at 2:36
  • $\begingroup$ Thanks again! Would this solution work for rotated ellipses aswell? If not, is it easy to adapt? $\endgroup$ – as1984 Jun 17 '16 at 2:42
  • $\begingroup$ The distance does not change as you rotate the ellipse. Distances do not change under rotation. Standard position made it easy to find the $x$ coordinate of the point. The formula, which only depends on $a$ and $b$, applies to all ellipses. $\endgroup$ – Ross Millikan Jun 17 '16 at 3:51

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