Eliminating the parameter from the parametric equations $x=h+a \sec t$ and $y=k+b \tan t$ gives the equation $\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$, which is a hyperbola. Find a set of parametric equations for a hyperbola with vertices at $(0,8)$ and $(0,-8)$ and foci at $(0,10)$ and $(0,-10)$. \[ x=\square, y= \] (Simplify your answers.)

Step 1 ：We are given a hyperbola with vertices at (0,8) and (0,-8) and foci at (0,10) and (0,-10). The center of the hyperbola is at (0,0).

Step 2 ：The standard form of the equation of a hyperbola centered at (h, k) with vertices a units from the center and foci c units from the center is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) if the transverse axis is horizontal and \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) if the transverse axis is vertical.

Step 3 ：In this case, the hyperbola is centered at (0, 0) with vertices at (0, 8) and (0, -8) and foci at (0, 10) and (0, -10). This means that the transverse axis is vertical, so we use the second form of the equation.

Step 4 ：The distance from the center to a vertex is a, so a = 8. The distance from the center to a focus is c, so c = 10. We can find b using the relationship \(c^2 = a^2 + b^2\).

Step 5 ：Substituting the values of a and c into the equation, we get \(b = \sqrt{c^2 - a^2} = \sqrt{10^2 - 8^2} = 6\).

Step 6 ：The parametric equations for a hyperbola in this form are \(x = h + b \sinh t\) and \(y = k + a \cosh t\). Substituting the values of h, k, a, and b into these equations, we get the parametric equations for the hyperbola as \(x = 6\sinh t\) and \(y = 8\cosh t\).

Step 7 ：\(\boxed{x = 6\sinh t, y = 8\cosh t}\) is the final answer.