Final answer:
The equation of an ellipse with vertices at (2,9) and (2,-5), and foci at (2,5) and (2,-1) is \( \frac{(x-2)^2}{40} + \frac{(y-2)^2}{49} = 1 \).
Step-by-step explanation:
To find the equation of an ellipse with vertices at (2,9) and (2,-5), and foci at (2,5) and (2,-1), we must first understand the properties of an ellipse. In an ellipse, the distance from any point on the ellipse to the two foci sums to a constant value. Vertices are the points where the ellipse intersects its major axis, which is the longest diameter of the ellipse.
The fact that the vertices and foci both have the same x-coordinate, 2, indicates the major axis is vertical, and thus our ellipse's equation will be in the form of \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) where (h,k) is the center of the ellipse, a is the semi-major axis, b is the semi-minor axis, and a > b.
The distance between the vertices is the length of the major axis, which is 9 - (-5) = 14, so the semi-major axis a = 14/2 = 7. The distance between the foci is 5 - (-1) = 6, so the focal length 2c = 6, giving us c = 3. Using the formula c^2 = a^2 - b^2, we get b^2 = a^2 - c^2 = 7^2 - 3^2 = 49 - 9 = 40.
The center of the ellipse is the midpoint between the vertices, which is (2, (9-5)/2) or (2,2). Therefore, the equation of the ellipse is:
\( \frac{(x-2)^2}{40} + \frac{(y-2)^2}{49} = 1 \)