JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Hyperbola

For the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$: since $25>9$, the major axis is along $y$, with $a_e^2=25,b_e^2=9$, so


Hyperbola eccentricity: $e_H=\frac{15}{8}{e}_e=\frac{15}{8}\cdot \frac{4}{5}=\frac{3}{2}$.

The hyperbola shares these foci, so it has the form $\frac{y^2}{A^2}-\frac{x^2}{B^2}=1$ (transverse axis along $y$, same as the foci), with $A{e}_H=4\Rightarrow A=\frac{8}{3}$, and


(One can check the given point $P\left(\sqrt{2},\frac{14}{3}\sqrt{\frac{2}{5}}\right)$ satisfies $\frac{y^2}{64/9}-\frac{x^2}{80/9}=1$ exactly.)

For a point $(x_0,y_0)$ on this hyperbola, the distance to the focus $(0,4)$ nearer the origin is $r=|e_H{y}_0-A|$. With $y_0=\frac{14}{3}\sqrt{\frac{2}{5}}$:


Since $7\sqrt{2/5}>A=8/3\approx 2.67$, this focal distance is


which is smaller than the other focal distance ($r+2A$), so this is the smaller focal distance.