Here foci are (±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 (±5, 0) ⇒ ae = 5, Length of transverse axis, 2a = 8 ⇒ a = 4 Now, ae = 5 ⇒ e = $\frac{5}{a}$ ⇒ e = $\frac{5}{4}$ We know that b = $a\sqrt{e^2-1}$ ⇒ b = $4\sqrt{\frac{25}{16}-1}$ = 4 × $\frac{3}{4}$ = 3 Thus required equation of hyperbola is $\frac{x^2}{4^2}-\frac{y^2}{3^2}$ = 1 ⇒ $\frac{x^2}{16}-\frac{y^2}{9}$ = 1.