# Thread: Finding the vertices, center and foci for an ellipse

1. ## Finding the vertices, center and foci for an ellipse

Center: ( , )
Right vertex: ( , )
Left vertex: ( , )
Top vertex: ( , )
Bottom vertex: ( , )
Right focus: ( , )
Left focus: ( , )

Rearranging this is giving me trouble for some reason. Any help is appreciated. Thanks in advance!

2. Since they've given you no information (no equation, no points, no graph, no description, etc), there's no way to find the answers they're wanting you to fill in. Sorry!

To learn how to work with ellipses and their equations, try here. But to answer this exercise, you'll first need to contact your instructor for the missing information.

3. Originally Posted by Beeorz
There must have been a transient error with the LSU server. The above image wasn't displaying earlier. In case it disappears again, the equation is:

. . . . . $9x^2\, +\, 16y^2\, -\, 90x\, -\, 256y\, +\, 1105\, =\, 0$

The first step, as always, is to complete the square to start converting the above to "conics" form.

. . . . . $9x^2\, -\, 90x\, +\, 16y^2\, -\, 256y\, =\, -1105$

. . . . . $9(x^2\, -\, 10x)\, +\, 16(y^2\, -\, 16y)\, =\, -1105$

. . . . . $9(x^2\, -\, 10x\, +\, 25)\, +\, 16(y^2\, -\, 16y\, +\, 64)\, =\, 9(25)\, +\, 16(64)\, -\, 1105$

. . . . . $9(x\, -\, 5)^2\, +\, 16(y\, -\, 8)^2\, =\, 144$

Divide through and simplify to get this into "conics" form.

Then use the fact that the larger denominator is under the numerator with an x in it to confirm that this ellipse is wider than it is tall.

Read the values off the "conics" form of the equation for h, k, a, and b.

Use the equation you've memorized to find the value of c.

Then fill in the various values for which they've asked you.

If you get stuck, please reply with a clear listing of your steps and reasoning so far. Thank you!

4. got it, was simply subtracting 9(25) and 16(64) instead of adding them to -1105

thanks for the help