Q. The eccentricity of the hyperbola in the standard form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, passing through (3, 0) and ( 3,$\sqrt{2}$, 2) is
A) $\frac{13}{3}$
B) $\sqrt{13}$
C) $\sqrt{3}$
D) $\frac{\sqrt{13}}{3}$
Explanation
(d)
Since, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,\ passes\ through\ (3,\ 0)and\ \left( 3\sqrt{2},2 \right)$
$\therefore\ \frac{9}{a^{2}} = 1\$
$\Rightarrow \ \ \ \ \ \ a^{2} = 9$
and $\frac{9 \times 2}{9} - \frac{4}{b^{2}} = 1\ \Rightarrow b^{2} = 4$
$\therefore\ e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{4}{9}} = \frac{\sqrt{13}}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.