Find the standard form of the equation of the hyperbola with the given characteristics. ​ Vertices: (6, 0),(10, 0); foci: (0, 0), (12, 0)

Answer 1

2a = 10 - 6 = 4; a = 2; 2c = 12 - 0 = 12, c = 6, b² = c² - a² = 36 - 4 = 32.Center at point ((0 + 12)/2, (0 + 0)/2) = (6, 0).Equation of hyperbola: (x - 6)²/4 - y²/32 = 1