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The equation of an ellipse, whose vertices are $(2, -2), (2, 4)$ and eccentricity $\frac{1}{3}$, is
$\frac{{{{(x - 2)}^2}}}{9} + \frac{{{{(y - 1)}^2}}}{8} = 1$
$\frac{{{{(x - 2)}^2}}}{8} + \frac{{{{(y - 1)}^2}}}{9} = 1$
$\frac{{{{(x + 2)}^2}}}{8} + \frac{{{{(y + 1)}^2}}}{9} = 1$
$\frac{{{{(x - 2)}^2}}}{9} + \frac{{{{(y + 1)}^2}}}{8} = 1$
Solution
(b) Given, co-ordinates of the vertices $(A$ and $A')$ of ellipse
$ \equiv (2,\, – 2)$ and $(2,\,4)$ and eccentricity $(e) = \frac{1}{3}$. We know that
$AA'$ is along a line parallel to y-axis. Therefore mid-point of
$AA'$ $(C) = (h,\,k) = (2,\,1)$ and distance between $AA'(2b) = 6$
or $b = 3$. We also know that the standard equation of an ellipse
at co-ordinates $(h,\,k)$ is $\frac{{{{(x – h)}^2}}}{{{a^2}}} + \frac{{{{(y – k)}^2}}}{{{b^2}}} = 1$ and
${a^2} = {b^2}(1 – {e^2}) = 9\left( {1 – \frac{1}{9}} \right) = 8.$ Therefore equation of the
ellipse $\frac{{{{(x – 2)}^2}}}{8} + \frac{{{{(y – 1)}^2}}}{9} = 1.$
