Analytic Geometry – Problem 28: Find the minimum possible value of

Let $E$ be the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ with foci $F_1(-3,0)$, $F_2(3,0)$. Let $M$ be any point on the circle $(x+3)^2+y^2=1$, and $N$ be any point on the circle $(x-3)^2+y^2=4$.

Point $P$ moves on $E$. Find the minimum possible value of $|PM|+|PN|$.

(1) Note that the two circles are centered at the foci: $MF_1=1$ and $NF_2=2$.

(2) By the triangle inequality, $$|PM|\ge |PF_1|-|MF_1|=|PF_1|-1,\qquad |PN|\ge |PF_2|-|NF_2|=|PF_2|-2.$$ Adding gives $$|PM|+|PN|\ge (|PF_1|+|PF_2|)-3.$$ (3) For $P\in E$, $|PF_1|+|PF_2|=2a=10$. Therefore $$|PM|+|PN|\ge 10-3=7.$$ (4) Equality can be achieved by choosing $P\in E$ and taking $M$ on segment $PF_1$ with $MF_1=1$, and $N$ on segment $PF_2$ with $NF_2=2$. Then $|PM|=|PF_1|-1$, $|PN|=|PF_2|-2$, so the lower bound $7$ is attainable. \]

The minimum value of $|PM|+|PN|$ is $7$.