Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: $(-2,0),(2,0) ; y$-intercepts: -8 and 8

Step 1 ：The standard form of the equation of an ellipse with center at the origin is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The distance between the foci is \(2c\), where \(c = \sqrt{a^2 - b^2}\). The y-intercepts are at \((0, \pm b)\).

Step 2 ：Given that the foci are at \((-2,0)\) and \((2,0)\), we know that \(2c = 4\), so \(c = 2\).

Step 3 ：Given that the y-intercepts are at \(-8\) and \(8\), we know that \(b = 8\).

Step 4 ：We can use these values to solve for \(a\).

Step 5 ：Using the values of \(c\) and \(b\), we find that \(a = \sqrt{c^2 + b^2} = \sqrt{4 + 64} = \sqrt{68} = 8.246211251235321\).

Step 6 ：Now that we have the values of \(a\), \(b\), and \(c\), we can substitute them into the standard form of the equation of an ellipse to get the final equation.

Step 7 ：Substituting \(a = \sqrt{68}\), \(b = 8\), and \(c = 2\) into the standard form of the equation of an ellipse, we get \(\frac{x^2}{68} + \frac{y^2}{64} = 1\).

Step 8 ：Final Answer: The standard form of the equation of the ellipse satisfying the given conditions is \(\boxed{\frac{x^2}{68} + \frac{y^2}{64} = 1}\).