Since you know that the points are colinear and you know their distances from the midpoint, a simple way to find the vertices is to compute the vectors from the midpoint to the foci and scale them to have the right length. You’ve got the midpoint $(5,2)$, so the two vectors are $(2,6)-(5,2)=(-3,4)$ and its negative. The $c$ value is $5$, so these vectors have length $5$ (you can check that for yourself). You need vectors of length $9$: scale them appropriately and add them to the midpoint to get the two vertices.
Let $B$ be the apex of a right circular cone. Let $\beta$ be the angle made between a generatrix of the cone and its axis. Let $\mathscr{P}$ be a plane not containing $B$ and intersecting the cone at an angle $\alpha$ with the axis. Let $\mathscr{Q}$ be the plane containing the axis and perpendicular to $\mathscr{P}$.
Construct the Dandelin sphere tangent to $\mathscr{P}$ and each generatrix of the cone. If $\alpha>\beta$, its center will be the incenter of the triangle cut in $\mathscr{Q}$ by $\mathscr{P}$ and the cone; if $\alpha<\beta$, it will be the excenter. (The case $\alpha=\beta$ can be handled as the excenter.)
Let $F$ be the point of tangency $\mathscr{P}$ with the sphere. Let $\mathscr{R}$ be the plane containing the circle of intersection between the sphere and the cone. Let $m$ be its line of intersection with $\mathscr{P}$. I claim that the section cut in the cone by $\mathscr{P}$ is the conic with focus $F$, directrix $m$, and eccentricity $\varepsilon=\frac{\cos\alpha}{\cos\beta}$.
To see this, let $A$ be a point on the section. Let $C$ be the foot of the perpendicular from $A$ to $m$. Let $D$ be the foot of the perpendicular from $A$ to $\mathscr{R}$, and let $E$ be the point where the generatrix through $A$ intersects the circle of intersection.
Let $x = AD$. Note that the measure of $\angle DAE$ is $\beta$ and that the measure of $\angle DAC$ is $\alpha$. Because $\overline{AF}$ and $\overline{AE}$ are both tangent to the sphere, $AE=AF$. So $\cos\beta=\frac{x}{AF}$. On the other hand, $\cos\alpha=\frac{x}{AC}$. So $\frac{\cos\alpha}{\cos\beta}=\frac{AF}{AC}$.
It follows that ratio of $AF$ to the distance from $A$ to $m$ is constant, and equal to $\varepsilon=\frac{\cos\alpha}{\cos\beta}$. So $A$ is on the conic with focus $F$, directrix $m$, and eccentricity $\varepsilon=\frac{\cos\alpha}{\cos\beta}$.
Best Answer
Like you have mentioned it could be the distance between two curves. However, here its the distance between any two points on a single curve. Consider the function $f(t) = t - \frac{1}{t}$. So here, they are basically asking the distance between points $(x,f(x))$ and $(y,f(y))$.
Considering $y = x - \frac{1}{x}$,we can rearrange it as $x(x-y)=1$. Thats simply a hyperbola. In the question, its mentioned one point has to be taken on the positive x axis and the other on the negative side. Hence we have to find the shortest distance between two branches of the hyperbola and thats the distance between its two vertices.