e and $e^{1}$ are the eccentricities of the hyperbolas $16 x^{2} - 9 y^{2} = 144$ and $9 x^{2} - 16 y^{2}$ = - 144 then e - $e^{1}$ =

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$\frac{3}{2}$

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Solution

The correct option is C
0

$\begin{matrix} 16 x^{2} - 9 y^{2} = 144 & 9 x^{2} - 16 y^{2} = - 144 \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 & \frac{x^{2}}{16} - \frac{y^{2}}{9} = - 1 e = \sqrt{1 + \frac{16}{9}} & e^{1} = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} & \frac{5}{3} \end{matrix}$

= e = $e^{1}$ = 0

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Similar questions

e and $e^{1}$ are the eccentricities of the hyperbolas $16 x^{2} - 9 y^{2} = 144$ and $9 x^{2} - 16 y^{2}$ = - 144 then e - $e^{1}$ =

e_{1}

e_{2}

16 x^{2} + 9 y^{2} = 144

16 x^{2} - 9 y^{2} = 144

$write the eccentricity of hyperbola$

$9 x^{2} - 16 y^{2} = 144$

16 x^{2} - 9 y^{2} = 144.

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