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

The correct option is B

$e_1^2+e_2^2=4$

Finding the value of the given expression:

Given equation can be rewritten as $x^2–y^2=\frac{25}{3}$

Here, $a^2=1,b^2=1$

$\begin{array}{rcl}\therefore e_1& =& 1+\sqrt{\frac{b^2}{a^2}}\\ & =& \sqrt{1+1}\\ & =& \sqrt{2}\end{array}$

The equation of the conjugate hyperbola is $–x^2+y^2=\frac{25}{3}$

$\begin{array}{rcl}\therefore e_2& =& \sqrt{1+\frac{a^2}{b^2}}\\ & =& \sqrt{1+1}\\ & =& \sqrt{2}\\ \therefore e_1^2+e_2^2& =& {\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2\\ & =& 2+2\\ & =& 4\end{array}$

Hence, the correct answer is option (B).