Find the equation of the hyperbola whose vertices are (±6,0) and one of the directrices is x = 4.

As the vertices are on the x-axis and their middle point is the origin, the equation is of type x²/a² - y²/b² = 1

Here b² = a²(e² -1), vertices are (±a,0) and directrices are given by x = ± a/e

Thus a = 6, a/e = 4 ⇒ e = 3/2 

⇒ b² = 36(9/4 - 1) = 45

∴ The required equation of the hyperbola is x²/36 - y²/45 = 1