If $e_{1\ }$is the eccentricity of the ellipse $\frac{x^{2}}{16}$ +$\frac{y^{2}}{7}$ = 1 and $e_{2}$ is the eccentricity of the hyperbola $\frac{x^{2}}{9} - \ \ \frac{y^{2}}{7}$ $= 1$, then $e_{1} + \ e_{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Given, ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$
$	herefore\ \ e_{1} = \sqrt{1 - \frac{7}{16}} = \frac{3}{4}$
and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{7} = 1$
$	herefore\ \ e_{2} = \sqrt{1 + \frac{7}{9}} = \frac{4}{3}$
Now, $e_{1} + e_{2} = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}$

Q. If $e_{1\ }$is the eccentricity of the ellipse $\frac{x^{2}}{16}$ +$\frac{y^{2}}{7}$ = 1 and $e_{2}$ is the eccentricity of the hyperbola $\frac{x^{2}}{9} - \ \ \frac{y^{2}}{7}$ $= 1$, then $e_{1} + \ e_{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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