Find the equation of an ellipse satisfying the given conditions. Foci: $(-4,0)$ and $(4,0)$ : length of major axis: 14

Step 1 ：The equation of an ellipse in standard form is given by \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h,k)\) is the center of the ellipse, \(a\) is the semi-major axis length, and \(b\) is the semi-minor axis length.

Step 2 ：The foci of the ellipse are located at \((h \pm c, k)\), where \(c\) is the distance from the center to each focus.

Step 3 ：Given that the foci are at \((-4,0)\) and \((4,0)\), we can determine that the center of the ellipse is at the midpoint of the line segment connecting the foci, which is \((0,0)\).

Step 4 ：The distance between the foci is \(2c\), so \(2c = 4 - (-4) = 8\). Therefore, \(c = 4\).

Step 5 ：The length of the major axis is given as 14, so \(2a = 14\) and \(a = 7\).

Step 6 ：We can find \(b\) using the relationship \(a^2 = b^2 + c^2\).

Step 7 ：Let's calculate \(b\) and write the equation of the ellipse.

Step 8 ：\(a = 7\)

Step 9 ：\(c = 4\)

Step 10 ：\(b = \sqrt{a^2 - c^2} = \sqrt{49 - 16} = \sqrt{33}\)

Step 11 ：The semi-minor axis length, \(b\), is approximately 5.74. Therefore, the equation of the ellipse is \(\frac{x^2}{49} + \frac{y^2}{33} = 1\).

Step 12 ：Final Answer: The equation of the ellipse is \(\boxed{\frac{x^2}{49} + \frac{y^2}{33} = 1}\).