Find the standard form of the equation of the ellipse with the given characteristics. Vertices: \( (2,2),(2,14) ; \) minor axis of length 4

To find the standard form of the equation of the ellipse, first, we identify the center and the lengths of the semi-major and semi-minor axes. The vertices are given as \( (2,2) \) and \( (2,14) \). The center of the ellipse is the midpoint of these vertices, calculated as follows: \[ \text{Center} = \left( \frac{2+2}{2}, \frac{2+14}{2} \right) = (2, 8) \] The distance from the center to either vertex gives us the semi-major axis length. The semi-major axis \( a \) is: \[ a = \frac{14 - 2}{2} = 6 \] Since the minor axis has a length of 4, the semi-minor axis \( b \) is: \[ b = \frac{4}{2} = 2 \] The ellipse is vertical since the major axis aligns with the y-axis. Therefore, the standard form of the equation of the ellipse is: \[ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \] Substituting \( h = 2, k = 8, a = 6, \) and \( b = 2 \), we have: \[ \frac{(x-2)^2}{2^2} + \frac{(y-8)^2}{6^2} = 1 \] This simplifies to: \[ \frac{(x-2)^2}{4} + \frac{(y-8)^2}{36} = 1 \] So the standard form of the equation of the ellipse is: \[ \frac{(x-2)^2}{4} + \frac{(y-8)^2}{36} = 1 \]