equation that has its foci at (-2,-3-2 square root 3)(-2, 3+ 2 square root 3) conjugate axis 6 units /
F1 = (-2, -3 - 2rt(3))
F2 = (-2, -3 + 2rt(3))
2F = distance between the focal points = 2*(2rt(3)) = 4rt(3) (difference between y-coordinates)
and F = rt(a^2 - b^2) (definition of distance between origin and a focal point)
or F^2 = a^2 - b^2
and b is the minor (or conjugate) axis length. i.e. b = 6.
Therefore, [2rt(3)]^2 = a^2 - 6^2
4*3 = a^2 - 36
a^2 = 48
a = 4rt(3)
The coordinate points for the foci are: (-2, -3 - 2rt(3)), (-2, -3 + 2rt(3)).
The both have the same x-coordinate, so the (ellipse) origin will also have the same x=coordinate, i.e. x = -2.
The y-coordinate of the origin will be halfway between the two foci y-coordinates,
i.e. (y1 + y2)/2 = (-3 - 2rt(3) - 3 + 2rt(3))/ 2 = (-6)/2 = -3
So, the origin of the ellipse is (xo, yo) = (-2, -3)
Ordinarily, the equation of an ellipse is defined by: (x - xo)^2/a^2 + (y - yo)^2 /b^2 = 1.
However, foci always lie on the major (longest) axis. And we have here foci that are lying vertically, therefore the major axis should be vertical, which means that the equation of the ellipse now should be,
(x - xo)^2/b^2 + (y - yo)^2 /a^2 = 1.
(x + 2)^2/36 + (y+3)^2/48 = 1