Question

If the foci of a hyperbola are the 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 \( \left( \sqrt{2}, \frac{14}{3} \sqrt{5} \right) \) on the hyperbola, is equal to

• A. \( 7 \sqrt{\frac{2}{5}} - \frac{8}{3} \)
• B. \( 14 \sqrt{\frac{2}{5}} - \frac{4}{3} \)
• C. \( 14 \sqrt{\frac{2}{5}} - \frac{16}{3} \)
• D. \( 7 \sqrt{\frac{2}{5}} + \frac{8}{3} \)

Answer 1

The Correct Option is A

Given:

\[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \] \[ a = 3, \; b = 5 \] \[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \quad \therefore \text{foci} = (0, \pm be) = (0, \pm 4) \] \[ e_1 = \frac{4}{5} \times \frac{15}{8} = \frac{3}{2} \]

Let equation hyperbola

\[ \frac{x^2}{A^2} - \frac{y^2}{B^2} = -1 \] \[ \therefore B = e_1 = 4 \quad \therefore B = \frac{8}{3} \] \[ \therefore A^2 = B^2 \left( e_1^2 - 1 \right) = \frac{64}{9} \left( \frac{9}{4} - 1 \right) \quad \therefore A^2 = \frac{80}{9} \]

\[ \frac{x^2}{80} - \frac{y^2}{64} = -1 \]

Directrix:

\[ y = \pm \frac{B}{e_1} = \pm \frac{16}{9} \] \[ PS = e \cdot PM = \frac{3}{2} \left[ \frac{14}{3} \cdot \sqrt{\frac{2}{5} - \frac{16}{9}} \right] \] \[ = 7 {\sqrt\frac{2}{5} - \frac{8}{3}} \]