alainjulius wrote:Can't find solutions on any modules.

They've got ellipses done

here.

alainjulius wrote:Find the equation of the ellipse given directrix at y = (9)(sqrt of 5)/5 + 2, center at (2, -6) and one

of the vertices on (2, -4).

I did the graph. That's a good place to start. From the center and the one vertex, you know that the other vertex is at (2, -8), so the long axis is 4 units across and "a" is 2. The

directrix for an ellipse is some constant-ratio (straight-line) distance from the nearby focal point as the focal point is (diagonally-ish) from the ellipse. (It's hard to say in words. The picture in that link is helpful.) The nearby focus is (2, c). That site in the second link says that the relationship is y = a^2/c = 4/c. So let's solve for c:

9sqrt[5]/5 + 2 = 4/c

9sqrt[5]/5 = 4/c - 2 = (4 - 2c)/c

c/5 = (4 - 2c)/9sqrt[5]

c^2/25 = (16 - 16c + 4c^2)/81*5

c^2/5 = (16 - 16c + 4c^2)/81

81c^2 = 80 - 80c + 20c^2

61c^2 + 80c - 80 = 0

Then do the Quadratic Formula to get c. Then you can get the whole equation, because you have h, k, a, and c. If you get stuck, please write back. Thanks!