EQUATION OF ELLIPSE given directrix at y = ...

[alainjulius]

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).

Spent hours on this one. I don't know if this is possible to solve. Can't find solutions on any modules. Please help. Thank you so much.

[maggiemagnet]

Can't find solutions on any modules.

They've got ellipses done here.

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!

[alainjulius]

Hey, thanks for the reply. I just wanna ask how did you arrive on the equation 9(sqrt{5})/5 + 2 = 4/c?Also, I tried to solve for the c without actually looking at your solution and I didn't get to arrive at solving the c using quadratic formula. I cross-multiplied c to the left side so I end up with 4/9(sqrt[5])/5 and I solve for the c.I think it's algrebraically correct.And if just in case I'm wrong, what value would you choose between two C's, cause accdng to your solution we have two C values, right? thank you.

[alainjulius]

ah I already understand the equation. I just wanna clarify your solution for arriving at the quadratic equation. thanks