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
$-x^2+y^2=\frac{25}{3}$
$\therefore e_2=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+1}=\sqrt{2}$
$\therefore e_1^2+e_2^2=(\sqrt{2}{)}^2+(\sqrt{2}{)}^2=4$