Find the standard form of the equation of the ellipse satisfying the given condit Foci: $(-4,0),(4,0) ; y$-intercepts: -7 and 7 Type the standard form of the equation. (Type an equation. Simplify your answer.)

Step 1 ：The standard form of the equation of an ellipse with its center at the origin (0,0) and its foci along the x-axis is given by: \(x^2/a^2 + y^2/b^2 = 1\), where a is the distance from the center to a vertex (half the length of the major axis) and b is the distance from the center to a co-vertex (half the length of the minor axis).

Step 2 ：The distance between the foci is 2c, where c is the distance from the center to a focus. In this case, the foci are given as (-4,0) and (4,0), so c = 4.

Step 3 ：The y-intercepts are given as -7 and 7, so the length of the minor axis is 14, and therefore b = 7.

Step 4 ：We can find a using the relationship \(a^2 = b^2 + c^2\).

Step 5 ：Let's calculate a: \(a = \sqrt{b^2 + c^2} = \sqrt{7^2 + 4^2} = 8.06225774829855\).

Step 6 ：Now that we have the values of a, b, and c, we can write the equation of the ellipse in standard form.

Step 7 ：Substitute a, b, and c into the equation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we get \(\frac{x^2}{8.06225774829855^2} + \frac{y^2}{7^2} = 1\).

Step 8 ：Simplify the equation to get the final answer: \(\boxed{\frac{x^2}{65.0} + \frac{y^2}{49} = 1}\).