Question

The foci of the hyperbola 2x² − 3y² = 5 are
(a) $(\pm 5/\sqrt{6},0)$
(b) (± 5/6, 0)
(c) $(\pm \sqrt{5}/6,0)$
(d) none of these

Answer 1

(a) $(\pm 5/\sqrt{6},0)$

The given equation of hyperbola is $2x^2-3y^2=5$. It can be rewritten in the following way:
$$\begin{gathered} \frac{2x^2}{5}-\frac{3y^2}{5}=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \frac{5}{2}}}-\frac{y^2}{{\displaystyle \frac{5}{3}}}=1 \end{gathered}$$

This is the standard equation of a parabola, where $a^2=\frac{5}{2}\mathrm{and}{b}^2=\frac{5}{3}$.
The eccentricity can be calculated in the following way:
$$\begin{gathered} b^2=a^2\left(e^2-1\right) \\ \Rightarrow \frac{5}{3}=\frac{5}{2}\left(e^2-1\right) \\ \Rightarrow e^2-1=\frac{2}{3} \\ \Rightarrow e^2=\frac{5}{3} \\ \Rightarrow e=\sqrt{\frac{5}{3}} \end{gathered}$$
Coordinates of the foci = $\left(\pm ae,0\right)=\left(\pm \frac{5}{\sqrt{6}},0\right)$