jmc

The lengths of the semi-major and semi-minor axis are $\sqrt{25}=5$ and $\sqrt{9}=3.$ Then the distance from the center $(0,0)$ of the ellipse to each focus is $\sqrt{5^2-3^2}=4,$ so the foci have coordinates $(\pm 4,0).$

Without loss of generality, assume that the parabola has its focus at $(4,0).$ Its directrix is the line containing the minor axis, which is the $y-$axis. Then the vertex of the parabola must be the point $(2,0),$ so its equation is of the form $$x=A{y}^2+2$$for some value of $A.$ Since the distance from the vertex to the focus is $2,$ we have $2=\frac{1}{4A},$ so $A=\frac{1}{8},$ and the equation of the parabola is $$x=\frac{y^2}{8}+2.$$The parabola and ellipse are shown together below. To find the intersection points of the parabola and ellipse, we solve the system $$\begin{array}{rl}\frac{x^2}{25}+\frac{y^2}{9}& =1,\\ x& =\frac{y^2}{8}+2.\end{array}$$Multiplying the first equation by $9$ and the second by $8,$ we can then eliminate $y$ by adding the two equations: $$\frac{9x^2}{25}+y^2+8x=y^2+25,$$or $$9x^2+200x-625=0.$$This quadratic factors as $$(9x-25)(x+25)=0.$$Since $x=\frac{y^2}{8}+2,$ it must be positive, so we have $x=\frac{25}{9}.$ Solving for $y$ in the equation $\frac{25}{9}=\frac{y^2}{8}+2,$ we get $y=\pm \frac{2\sqrt{14}}{3}.$ Therefore, the distance between the two points is $2\cdot \frac{2\sqrt{14}}{3}=\boxed{\frac{4\sqrt{14}}{3}}.$