The minimal value of a fraction based on a focal chord of an ellipse

Question

I came across a very interesting olympiad problem. It goes as follows: suppose you have an ellipse given by $\frac{x^2}{16}+\frac{y^2}{9}=1$ and a line that goes through the point $A(\sqrt{7},0)$. This line also goes through the ellipse at two points $N$ and $M$. What is the minimum possible value of $\frac{AM\cdot AN}{NM}$? Here is a desmos link for visualization: https://www.desmos.com/calculator/mncpycyaj3 . My attempt: first I noted that $a=4$ and $b=3$ from the equation of the ellipse. Then I know that $a^2-b^2=c^2$, so $c=\sqrt{7}$. This immediately tells us that the line actually goes through the right focus $F_2$. So, the fraction of interest can be written as $\frac{AM\cdot AN}{NM}=\frac{F_2M\cdot F_2N}{F_2M+F_2N}$. Now I tried a few possible configurations of the line: 1) [a horizontal line] $N(4,0);M(-4,0)$, thus, $F_2N=\sqrt{(4-\sqrt{7})^2+(0-0)^2}=4-\sqrt{7}$ and $F_2M=\sqrt{(-4-\sqrt{7})^2+(0-0)^2}=4+\sqrt{7}$. So, $\frac{F_2M\cdot F_2N}{F_2M+F_2N}=\frac{(4-\sqrt{7})(4+\sqrt{7})}{8}=\frac{16-7}{8}=\frac{9}{8}=1.125$. 2) [a vertical line] $N\left(\sqrt{7},\frac{9}{4}\right);M\left(\sqrt{7},-\frac{9}{4}\right): F_2N=\sqrt{(\sqrt{7}-\sqrt{7})^2+\left(\frac{9}{4}-0\right)^2}=\frac{9}{4}=2.25=F_2M$, thus, $\frac{F_2M\cdot F_2N}{F_2M+F_2N}=\frac{2.25\cdot 2.25}{4.5}=1.125$. At this point I am thinking that this is some universal constant or something, but I have only checked the extreme cases. 3) [a line going through the highest point of the ellipse] $N(0,3)$, then the other point is found by $$y=\frac{y_M}{x_M-\sqrt{7}}(x-\sqrt{7});\space 3=-\frac{y_M}{x_M-\sqrt{7}}\sqrt{7}$$ But this point lies on the ellipse, $\frac{x_M^2}{16}+\frac{y_M^2}{9}=1$. Solving for $x_M$ and then for $y_M$ we get $M\left(\frac{32\sqrt{7}}{23},-\frac{27}{23}\right)$. Then $F_2N=\sqrt{(0-\sqrt{7})^2+(3-0)^2}=4$ and $F_2M=\sqrt{\left(\frac{32\sqrt{7}}{23}-\sqrt{7}\right)^2+\left(-\frac{27}{23}-0\right)^2}=\frac{36}{23}$;$\space \frac{F_2M\cdot F_2N}{F_2M+F_2N}=1.125$. My question is: is this fraction always equal to $1.125$ and if yes, then how can I prove this fact? I have tried some arbitrary points from desmos and plugged them in Wolframalpha, the outputs seem to be slightly bigger than 1.125, but I guess that is just due to rounding up of the points in desmos. And another thing I would like to know is: if it is indeed true then is this only a property of lines going through focuses or in general for any lines? Any hints are much appreciated.

Answer 1

This is a focal chord, so the product of the distances $AN$ and $AM$ is a constant

$AN \cdot AM = (a - c) \cdot (a + c) = a^2 - c^2 = b^2$

Thus

$AN \cdot AM = 9 $

The ratio requested is thus $\dfrac{ 9 }{ NM } $

and the distance ${NM}$ is maximum when $N$ and $M$ are on the major axis, in which case ${NM} = 2a = 8$

Thus the minimum possible ratio is $\dfrac{9}{8} = 1.125$