22. Vertices: \( (3,1),(3,11) \); minor axis of length 2

To find the equation of the ellipse, we start with the provided vertices. The vertices, \( (3,1) \) and \( (3,11) \), are aligned vertically, indicating that the center of the ellipse is at the midpoint of these vertices. Thus, the center is: \[ \left(3, \frac{1+11}{2}\right) = (3, 6) \] The distance between the vertices is the length of the major axis. The distance from \( (3,1) \) to \( (3,11) \) is: \[ 11 - 1 = 10 \] Hence, the semi-major axis \( a \) is half of this distance: \[ a = \frac{10}{2} = 5 \] The minor axis is given to be of length 2, so the semi-minor axis \( b \) is: \[ b = \frac{2}{2} = 1 \] Since the major axis is vertical, the standard form of the ellipse's equation is: \[ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \] Substituting in \( h = 3 \), \( k = 6 \), \( b = 1 \), and \( a = 5 \): \[ \frac{(x-3)^2}{1^2} + \frac{(y-6)^2}{5^2} = 1 \] This simplifies to: \[ \frac{(x-3)^2}{1} + \frac{(y-6)^2}{25} = 1 \] So, the equation of the ellipse is: \[ \frac{(x-3)^2}{1} + \frac{(y-6)^2}{25} = 1 \]