.
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.
As a prerequisite, see the lesson
- Ellipse definition, canonical equation, characteristic points and elements
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic
"Conic sections: Ellipses. Definition, major elements and properties. Solved problems".