Back Question 3 Next Question 3 of 5 (20 points) Find the equation of the hyperbola centered at the origin with vertices (16,0)and (−16,0)and asymptotes y=±14x.

Answer 1

Equation of hyperbola x²/a² - y²/b² = 1; a = 16 and b/a = 14, b = 14a = 14·16 = 224;

so equation of hyperbola: x²/16² - y²/224² = 1