Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form is $\frac{x^2}{a^{-y^2} b^2}$ = 1 ∴ Vertices are (±2, 0) ⇒ a = 2; Foci are (±3, 0) ⇒ ae = 3 Now, ae = 3 ⇒ e = $\frac{3}{a}$ ⇒ e = $\frac{3}{2}$ We know that b = $a\sqrt{e^2-1}$ ⇒ b = $2\sqrt{\frac{9}{4}-1}$ = 2 $\frac{\sqrt{5}}{2}$ = $\sqrt{5}$ Thus required equation of hyperbola is $\frac{x^2}{(2)^2} - \frac{y^2}{(\sqrt{5})^2}$ = 1 ⇒ $x^2 - \frac{4y^2}{5}$ = 1