If e₁ is the eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ and e₂ is the eccentricity of the hyperbola passing through the foci of the ellipse and e₁ e₂ = 1 then the equation of the hyperbola is

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$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = - 1$

$\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1$

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

answer is B.

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Detailed Solution

$G i v e n e l l i p s e \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 (w h e r e a < b) N o w e_{1} = \sqrt{\frac{9}{25}} = \frac{3}{5} f o c i = (0, \pm 5 (\frac{3}{5})) = (0, \pm 3) L e t t h e h y p e r b o l a b e \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1 \to (1) G i v e n t h a t e_{1} e_{2} = 1 \Rightarrow e_{2} = \frac{5}{3} (1) p a s s e s t h r o u g h (0, \pm 3) t h e n b^{2} = 9 a n d a^{2} = b^{2} (e^{2} - 1) = 9 (\frac{16}{9}) = 16 S o, h y p e r b o l a i s \frac{x^{2}}{16} - \frac{y^{2}}{9} = - 1$