The two vertices (and thus the two foci) lie on the y-axis. Consequently we are looking for an equation of the form, (y²/a²) - (x²/b²) = 1 Note : If the vertices had been on the x -axis then the equation would have been, (x²/a² ) - (y²/b²) = 1 As the vertices are at (0, ±10) then a = 10. The standard equation for the asymptotes is y = ±(a/b)x Note : If the vertices had been on the x - axis, then the equations would be y = ±(b/a)x We are told that y = ±(5/4)x. As a = 10 then we can generate an equivalent formula for the asymptotes of y = ±(10/8)x......which tells us that b = 8. Hence the equation of the hyperbola is (y²/10²) - (x²/8²) = 1.
