# Longest distance to the foci or the center that a point within the ellipse can be?

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!

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.
• 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. – Ross Millikan Jun 17 '16 at 2:36
• 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. – Ross Millikan Jun 17 '16 at 3:51