Here foci are (±3 $\sqrt{5}$,0) which lie on x-axis. So the equation of the hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1. Now, Foci are (±3 $\sqrt{5}$,0) ⇒ c = $3\sqrt{5}$. Length of latus rectum is $\frac{2b^2}{a}$ = 8 ⇒ $b^2$ = 4a We know that $c^2$ = $a^2 + b^2$ ∴ $(3\sqrt{5})^2$ = $a^2$ + 4a ⇒ $a^2$ + 4a – 45 = 0 ⇒ (a + 9)(a – 5) = 0 ⇒ a = 5 (Since a = –9 is not possible) ∴ $a^2$ = 25 and $b^2$ = 4 × 5 = 20 Thus required equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}$ = 1.