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My guess is this means $\displaystyle \frac{(x-2)^2}{36}+\frac{(y+3)^2}{16}$. This is still not an equation, but $\displaystyle \left(\frac{x-2}{6}\right)^2+\left(\frac{y+3}{4}\right)^2=1$ corresponds to an ellipse with the center at (2, -3).

By "vertices" do you mean foci (plural of "focus")? They are on the major axes at the distance $\displaystyle \sqrt{a^2-b^2}$ from the center (a and b are semi-major axis and semi-minor axis, respectively).

The vertices of an ellipse are not the foci, they are the points on the ellipse where the curvature is maximum or minimum. They are the points at the ends of the axes. For an ellipse of the form $\displaystyle \frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1$ they are at $\displaystyle (x_0, y_0+ b)$, $\displaystyle (x_0, y_0- b)$, $\displaystyle (x_0+ a, y_0)]$, and [tex]x_0- a, y_0)[]/tex].