The equation of an ellipse,whose vertices are | vedclass

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(B) The vertices are $(2, -2)$ and $(2, 4)$. The center $(h, k)$ is the midpoint of the vertices: $h = \frac{2+2}{2} = 2$ and $k = \frac{-2+4}{2} = 1$

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$\frac{(x - 2)^2}{8} + \frac{(y - 1)^2}{9} = 1$

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(B) The vertices are $(2, -2)$ and $(2, 4)$. The center $(h, k)$ is the midpoint of the vertices: $h = \frac{2+2}{2} = 2$ and $k = \frac{-2+4}{2} = 1$. The center is $(2, 1)$.The length of the major axis $2b = \sqrt{(2-2)^2 + (4 - (-2))^2} = 6$,so $b = 3$ and $b^2 = 9$.Since the vertices lie on a line parallel to the $y$-axis,the equation is of the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.Given eccentricity $e = \frac{1}{3}$,we use the relation $a^2 = b^2(1 - e^2) = 9(1 - \frac{1}{9}) = 9(\frac{8}{9}) = 8$.Substituting the values,the equation is $\frac{(x-2)^2}{8} + \frac{(y-1)^2}{9} = 1$.

The equation of an ellipse,whose vertices are $(2, -2)$ and $(2, 4)$ and eccentricity is $\frac{1}{3}$,is

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