Question

How do you find the foci and sketch the ellipse `x^2+9y^2=4`?

Answer 1

`(0, +-(4sqrt(2))/3)`

Explanation:

Ellipses always take the form of `((x-h)^2)/a^2` + `((y-k)^2)/b^2` = 1, where a and b are interchangeable, and a is always larger than b. So `((x-h)^2)/b^2` + `((y-k)^2)/a^2`= 1 is the other possibility.

The center of an ellipse is always at (h,k), and the equation is always equal to one. Because your equation is equal to 4, you must divide all the terms by 4 to get the ellipse into standard form:
`(x^2)/4` + `(9y^2)/4` = 1 `=>` `(x^2)/2^2 + (y^2/(2/3)^2)` = 1

a=2, and b =`2/3`, since 2 > `2/3`.

Your major axis is always length 2a. Your minor (smaller) axis is always 2b. Since the center of the ellipse is at (0,0), the vertices at either end of the major axis are at `(+-4,0)`. You know they're moving left and right from the center because they're part of the `x^2` term.

The ends of the minor axis are the "co-vertices", and they are distance b away from the center of the ellipse. Because b is under the y term, they are up and down from the center. So `(0, +-2/3)` are the endpoints of your minor axis.

Using these endpoints, you can sketch your ellipse.

Now, for the foci: For ellipses, remember that `c^2=a^2-b^2`. (Kind of like the opposite of Pythagoras' theorem), and c is the distance from the center to the foci.
`c^2=36/9 - 4/9 = 32/9`
so `c = (+-(4sqrt(2))/3)`

Foci always lie on the Major axis of an ellipse, so move left and right c from the origin to find your foci.

Hence, your foci are at `(0, +-(4sqrt(2))/3)`.