Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/7x.

A. y²/36 - x²/196 = 1
B. y²/196 - x²/36 = 1
C. y²/36 - x²/49 = 1
D. y²/49 - x²/9 = 1

Answer 1

The equation of the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/7x is y²/36 - x²/49 = 1.

Step-by-step explanation:

The equation of the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/7x can be written in the standard form y²/a² - x²/b² = 1. The value of 'a' is the distance from the center of the hyperbola to the vertices, which is 6. The value of 'b' can be found using the equation of the asymptotes: b/a = 3/7. Solving for 'b', we get b = (3/7)a. Substituting these values into the equation, we get the equation in standard form as: y²/36 - x²/49 = 1.