SOLUTION: Write an equation for the ellipse, satisfying the following conditions. Foci at (0,-2) and (0,2); the point (3,2) on the ellipse

1.  The center of the ellipse is at (0,0).

    The major axis        is y-axis x= 0;
    Hence, the minor axis is x-axis y= 0.


2.  The canonical equation of this ellipse is 

     = 1    (it is written in the form to fit the fact that the major axis is y-axis: a > b > 0)

    Since the point (3,2) is on the ellipse, it implies 

     = 1,   or

     = .    (1)


3.  The focal distance is 2c = 2 - (-2) = 4.
    Hence, the linear eccentricity c =  = 2. It means that

     = 4.          (2)


4.  Thus you have two equations (1) and (2) to determine "a" and "b".

    You can simplify writing and solving by introducing new variables x =  and y = :

    9x + 4y = xy         (3)     instead of (1), and
    x - y = 4.           (4)     instead of (2)

The setup is done.
Now it is simple arithmetic to solve it and to get "a" and "b" at the end.


4.  From (4), x = 4 + y, Substitute it into (3). You will get

    9(4+y) + 4y = (4+y)*y,

    36 + 9y + 4y = 4y + y^2  --->  y^2 -9y - 36 = 0  ---->   =  = .

    Only positive root works: y = 12.  So,  = 12 and b = .

    Then a^2 = b^2 + 4 = 12 + 4 = 16 and  a =  = 4.

    Thus semi-axes are  4 (vertical) and  (horizontal).


    The equation for the ellipse is

     +  = 1.