Final answer:
The standard form equation of the hyperbola with vertices at (8,0) and (-8,0) and an asymptote of y=(7/4)x is (x^2)/64 - (y^2)/196 = 1.
Step-by-step explanation:
The student has asked for the standard form equation of a hyperbola with vertices at (8,0) and (-8,0) and an asymptote of y = (7/4)x. To find the standard form of the hyperbola, we can use the fact that the vertices determine the distance of the center to the vertices (a), which is also the length of the semi-major axis. The slope of the asymptote is tied to the relationship between 'a' and 'b' (where b is related to the semi-minor axis), since the asymptotes of a hyperbola are given by y = ±(b/a)x when the hyperbola is centered at the origin.
Given that the vertices are on the x-axis and symmetric about the y-axis, the center of the hyperbola is (0, 0), and the hyperbola is horizontal. The distance from the center to a vertex is 8, so 'a' is 8. Now, because the slope of the asymptote is 7/4, and we know for the hyperbola the slope of the asymptote is 'b/a', we can set up the equation 7/4 = b/8, solving for 'b' gives us b = 14. So, the standard form for a horizontal hyperbola with center (0, 0) is:
(x^2)/(a^2) - (y^2)/(b^2) = 1
Plugging in our values for 'a' and 'b', we get the standard form equation for the hyperbola:
(x^2)/(8^2) - (y^2)/(14^2) = 1
Or simplifying:
(x^2)/64 - (y^2)/196 = 1