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