The foci of a hyperbola coincide with the foci of the ellipse, x216+y29=1. If the eccentricity of the hyperbola is 4, then its equation is

Eccentricity of ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 is \sqrt{\frac{16 - 9}{16}} = \frac{\sqrt{7}}{4}$

Foci of the ellipse $= (\pm \sqrt{7}, 0)$ . $∴$ Foci of the hyperbola $= (\pm \sqrt{7}, 0)$

Eccentricity of the hyperbola = 4.

$∴ ae = \sqrt{7} \Rightarrow 4 a = \sqrt{7} \Rightarrow a = \frac{\sqrt{7}}{4} . b^{2} = a^{2} e^{2} - a^{2} = 7 - \frac{7}{16} = \frac{105}{16}$

$∴$ Equation of the hyperbola is $\frac{x^{2}}{7 / 16} - \frac{y^{2}}{105 / 16} = 1 \Rightarrow \frac{x^{2}}{7} - \frac{y^{2}}{105} = \frac{1}{16}$ .

The foci of a hyperbola coincide with the foci of the ellipse, $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ . If the eccentricity of the hyperbola is 4, then its equation is