If the foci of a hyperbola are same as that of the ellipse $$\frac{x^2}{9} + \frac{y^2}{25} = 1$$ and the eccentricity of the hyperbola is $$\frac{15}{8}$$ times the eccentricity of the ellipse, then the smaller focal distance of the point $$\left(\sqrt{2}, \frac{14}{3}\sqrt{\frac{2}{5}}\right)$$ on the hyperbola, is equal to

Consider the ellipse given by $$\frac{x^2}{9} + \frac{y^2}{25} = 1.$$ Since the larger denominator is 25, the major axis runs along the y-axis, so we set $$a^2 = 25$$ and $$b^2 = 9.$$ It follows that $$c^2 = a^2 - b^2 = 25 - 9 = 16 \implies c = 4,$$ and the foci lie at $$(0,\pm4).$$ The eccentricity of this ellipse is $$e_1 = \frac{c}{a} = \frac{4}{5}.$$

The hyperbola we seek shares these foci at $$(0,\pm4)$$, so its transverse axis is also vertical and its focal parameter is $$c_h = 4.$$ We are told its eccentricity satisfies $$e_2 = \frac{15}{8}\,e_1 = \frac{15}{8}\times\frac{4}{5} = \frac{3}{2}.$$ Thus $$a_h = \frac{c_h}{e_2} = \frac{4}{3/2} = \frac{8}{3},$$ and $$b_h^2 = c_h^2 - a_h^2 = 16 - \frac{64}{9} = \frac{80}{9}.$$ Accordingly, the equation of the hyperbola is $$\frac{y^2}{64/9} \;-\;\frac{x^2}{80/9} \;=\;1.$$

To verify that the point $$P\bigl(\sqrt{2},\,\tfrac{14}{3}\sqrt{\tfrac{2}{5}}\bigr)$$ lies on this curve, note first that $$y^2 = \frac{196}{9}\times\frac{2}{5} = \frac{392}{45},$$ so $$\frac{y^2}{64/9} = \frac{392/45}{64/9} = \frac{392}{45}\times\frac{9}{64} = \frac{49}{40},$$ while $$\frac{x^2}{80/9} = \frac{2}{80/9} = \frac{9}{40}.$$ Their difference is $$\frac{49}{40} - \frac{9}{40} = 1,$$ confirming that $$P$$ indeed satisfies the hyperbola’s equation.

For any point on a hyperbola with eccentricity $$e_2$$ and transverse semi-axis $$a_h$$, the distances to the two foci satisfy $$|PF| = \bigl|e_2\,y \pm a_h\bigr|.$$ Since the y-coordinate of $$P$$ is positive, we compute $$e_2\,y = \frac{3}{2}\times\frac{14}{3}\sqrt{\frac{2}{5}} = 7\sqrt{\frac{2}{5}},$$ and hence the two focal distances are $$PF_1 = 7\sqrt{\frac{2}{5}} - \frac{8}{3},\qquad PF_2 = 7\sqrt{\frac{2}{5}} + \frac{8}{3}.$$ The smaller of these is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}.$$

Therefore, the smaller focal distance is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}$$.