Vertices are (0, ±3) which lie on y-axis. So the equation of hyperbola in standard form is $\frac{y^2}{a^2} - \frac{x^2}{b^2}$ = 1 Now vertices are (0, ±3) ⇒ a = 3; Co-ordinates of Foci are (0, ±5) ⇒ ae = 5 Now, ae = 5 ⇒ e = $\frac{5}{a}$ ⇒ e = $\frac{5}{3}$ We know that b = $a\sqrt{e^2-1}$ ⇒ b = $3\sqrt{\frac{25}{9} - 1}$ = $3\sqrt{\frac{16}{9}}$ = 4 Thus required equation of hyperbola is $\frac{y^2}{3^2} - \frac{x^2}{4^2}$ = 1 ⇒ $\frac{y^2}{9} - \frac{x^2}{16}$ = 1.