Find the standard form of the equation of the ellipse satisfying the given conditions.
Foci: $(-4,0),(4,0)$; vertices: $(-12,0),(12,0)$
Answer
Final Answer: The standard form of the equation of the ellipse satisfying the given conditions is \(\boxed{\frac{x^2}{144} + \frac{y^2}{128} = 1}\).
Step 1 ：Given the vertices and foci of the ellipse, we can find the values of a and c. The distance from the center to a vertex is \(a = 12\) and the distance from the center to a focus is \(c = 4\).
Step 2 ：We can use these values to find b using the relationship \(a^2 = b^2 + c^2\). After calculation, we find that \(b = \sqrt{a^2 - c^2} = \sqrt{12^2 - 4^2} = \sqrt{128} = 11.313708498984761\).
Step 3 ：Now that we have the values of a, b, and c, we can substitute these into the standard form of the equation of an ellipse to get the equation of the ellipse satisfying the given conditions.
Step 4 ：Substituting the values of a and b into the standard form of the equation of an ellipse, we get \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) becomes \(\frac{x^2}{144} + \frac{y^2}{128} = 1\).
Step 5 ：Final Answer: The standard form of the equation of the ellipse satisfying the given conditions is \(\boxed{\frac{x^2}{144} + \frac{y^2}{128} = 1}\).
