Find the standard form of the equation of the hyperbola satisfying the given conditions. Endpoints of transverse axis: $(0,-24),(0,24)$; asymptote: $y=2 x$

Step 1 ：The standard form of the equation of a hyperbola with a vertical transverse axis is given by \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where \((h,k)\) is the center of the hyperbola, \(2a\) is the length of the transverse axis, and \(2b\) is the length of the conjugate axis. The slopes of the asymptotes are \(\pm \frac{a}{b}\).

Step 2 ：From the given conditions, we can see that the center of the hyperbola is at \((0,0)\), and the length of the transverse axis is \(2a = 24 - (-24) = 48\), so \(a = 24\).

Step 3 ：The slope of the asymptote is \(2 = \frac{a}{b}\), so \(b = \frac{a}{2} = 12\).

Step 4 ：So, the standard form of the equation of the hyperbola is \(\frac{y^2}{24^2} - \frac{x^2}{12^2} = 1\).

Step 5 ：Final Answer: The standard form of the equation of the hyperbola is \(\boxed{-\frac{x^2}{144} + \frac{y^2}{576} = 1}\).