Question

If $e_1 $ and $e_2$ are the eccentricities of a hyperbola $3x^2 -3y^2 = 25$ and its conjugate, then

• A. $e_{1}^{2}+e_{2}^{2}=2$
• B. $e_{1}^{2}+e_{2}^{2}=4$
• C. $e_{1}+e_{2}=4$
• D. $e_1 +e_2 = \sqrt {2}$

Answer 1

Answer: (B)

Given equation can be rewritten as
$x^{2}-y^{2}=\frac{25}{3}$
Here, $a^{2}=1, b^{2}=1$
$\therefore \,\,\,e_{1}=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+1}=\sqrt{2}$
The equation of conjugate hyperbola is
$-x^{2}+y^{2}=\frac{25}{3}$
$\therefore \,\,\, e_{2}=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+1}=\sqrt{2}$
$\therefore \,\,\,\,e_{1}^{2}+e_{2}^{2}=(\sqrt{2})^{2}+(\sqrt{2})^{2}=4$