Hyperbola Question 6
Question 6 - 2024 (31 Jan Shift 1)
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
(1) $7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
(2) $14 \sqrt{\frac{2}{5}}-\frac{4}{3}$
(3) $14 \sqrt{\frac{2}{5}}-\frac{16}{3}$
(4) $7 \sqrt{\frac{2}{5}}+\frac{8}{3}$
Show Answer
Answer (1)
Solution
$\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$
$a=3, b=5$
$e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore$ foci $=(0, \pm b e)=(0, \pm 4)$
$\therefore e_{H}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}$
Let equation hyperbola
$\frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~B}^{2}}=-1$
$\therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3}$
$\therefore \mathrm{A}^{2}=\mathrm{B}^{2}\left(\mathrm{e}_{\mathrm{H}}^{2}-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^{2}=\frac{80}{9}$
$\therefore \frac{\mathrm{x}^{2}}{\frac{80}{9}}-\frac{\mathrm{y}^{2}}{\frac{64}{9}}=-1$
Directrix : $y= \pm \frac{B}{e_{\mathrm{H}}}= \pm \frac{16}{9}$
$\mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right|$
$=7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
