Question

How do you graph the ellipse `(x-4)^2/8+(y-2)^2/18=1` and find the center, the major and minor axis, vertices, foci and eccentricity?

Answer 1

Center: `4,2`
Major axis: length `3sqrt(2)`, vertical
Minor axis: legnth `2sqrt(2)`, horizontal
Vertices: `(2sqrt(2),0),(-2sqrt(2),0),(0,3sqrt(2)),(0,-3sqrt(2))`
Foci: `(0,-sqrt(10)),(0,sqrt(10))`
Eccentricity: `sqrt(5)/3`

Explanation:

Given the general formula

`(x-x_0)^2/(a^2) + (y-y_0)^2/(b^2) = 1`

you have that:
- `x_0` is the `x` component of the center
- `y_0` is the `y` component of the center
- `a` is the horizontal semi-axis
- `b` is the vertical semi-axis

This easily answers the question about center, axes and vertices.

For the rest, you have a formula for everything:

• Foci: `c = sqrt(b^2-a^2)` (in general, major axis minus minor axis)
• Eccentricity = `c/b` (in general, focus divided by major axis)