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

Let $$P$$ be a point on the hyperbola $$H: \frac{x^2}{9}-\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2 \sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is

Let $$e_1$$ be the eccentricity of the hyperbola $$\frac{x^2}{16}-\frac{y^2}{9}=1$$ and $$e_2$$ be the eccentricity of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a} > \mathrm{b}$$, which passes through the foci of the hyperbola. If $$\mathrm{e}_1 \mathrm{e}_2=1$$, then the length of the chord of the ellipse parallel to the $$x$$-axis and passing through $$(0,2)$$ is :