Hyperbola
Focal Distance — Hyperbola with Shared Foci
nta_pyq_2024_jan
Grade 11
Question:
If the foci of a hyperbola are same as that 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 equal to
$7\sqrt{\dfrac{2}{5}}-\dfrac{8}{3}$
$14\sqrt{\dfrac{2}{5}}-\dfrac{4}{3}$
$14\sqrt{\dfrac{2}{5}}-\dfrac{16}{3}$
$7\sqrt{\dfrac{2}{5}}+\dfrac{8}{3}$
Step-by-Step Solution
Key Concept: Ellipse $x^2/9+y^2/25=1$: $a=5,b=3$, $e_{ellipse}=4/5$, foci $(0,\pm4)$. Hyperbola has vertical foci $(0,\pm4)$: $Be_H=4$, $e_H=\frac{15}{8}\cdot\frac{4}{5}=\frac{3}{2}$, $B=8/3$. Then $A^2=B^2(e_H^2-1)$. Use focal distance formula $|e_H\cdot y-B/e_H|$ for the hyperbola.
Ellipse foci $(0,\pm4)$. $Be_H=4$, $e_H=3/2\Rightarrow B=8/3$. Directrix $y=\pm16/9$. Focal distance of point: $e_H\cdot|y_P-B/e_H|=\frac{3}{2}\left(\frac{14\sqrt{2/5}}{3}-\frac{16}{9}\right)=7\sqrt{\frac{2}{5}}-\frac{8}{3}$.
Correct Answer: 1