Final answer:
The equation of the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/7 is y²/36 - x²/(324/49) = 1.
Step-by-step explanation:
To find the equation of the hyperbola in standard form, you need to know two things: the coordinates of the vertices and the slopes of the asymptotes. Since the vertices are given at (0, ±6), we know the hyperbola is centered at the origin (0,0) and opens vertically. Additionally, the distance from the center to each vertex is 6 units, which represents the value of 'a' in the standard equation for a hyperbola.
The slopes of the asymptotes for a vertical hyperbola are ±b/a. Given that the slopes of the asymptotes are ±3/7, and we know 'a' is 6, we can determine 'b' by setting 3/7 equal to b/6. Solving for 'b' yields b = 18/7. Now we have all the components needed for the standard equation:
The Equation of a Hyperbola
Standard form for a vertical hyperbola: ±(y - k)²/a² - (x - h)²/b² = 1 where (h,k) is the center.
Substituting the known values into the equation, we get:
y²/6² - x²/(18/7)² = 1
Finally, simplifying the denominators, the equation becomes:
y²/36 - x²/(324/49) = 1
Standard Equation of the given hyperbola: y²/36 - x²/(324/49) = 1