Two problems have given me trouble..

9. Find the center, vertices, foci of x^2 over 25 + y^2 over 4 = 1

10. Find the center, vertices, foci of 9x^2 + 4y^2 - 18x + 16y - 11 = 0

If anyone knows how to do these please help

for 9, it is in standard form - use the standard form of an ellipse

for 10, complete the square to put it in standard form

if you cannot find the center of #9, you need to study more instead of posting here (not trying to be a smart ass)

shoul clarify how to put in standard form
he means (nx+a)*(my+b) where n, m, a, b are all constants.

The center is the number that is being added to x^2 or y^2. Opposite signs.

For example: (x^2 + 9) and (y^2 - 8) have the center of (-9, 8).

The vertices are determined by the positive and negative roots of the denominator. So
25 and 4 are the denominators so it's +5 and -5, +2 as well as -2.

Y = Rise
X = Run

For example: If the x denominator was 36, it'd be +6 and -6. If y denominator was 25, it'd
be +5 and -5. 6 units to the right/left of the center, and 5 units up/down of the center.

Erm I forgot foci

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Find the center, vertices, foci of 9x^2 + 4y^2 - 18x + 16y - 11 = 0

1. Put it in order

9x^2 - 18x + 4y^2 + 16y = 11

2. Pull out "A" for the "X" and "Y" portions

9(x^2 - 2x) + 4(y^2 + 4y) = 11

3. Divide "B" by 2 and square it. Take that number and add it
as "C."

9(x^2-2x + 1) + 4(y^2 + 4y + 4) = 11

4. Multiply the number you pulled out with "C" and add it
to the opposition.

9(x^2-2x + 1) + 4(y^2 + 4y + 4) = 11 + 9 + 16

5. Use perfect squares to complete it

9(x+1)^2 + 4(y+2)^2 = 36

6. The opposition has to be a 1, so divide everything by 36.

{[(x+1)^2]/4]} + {[(y+2)^2]/9} = 1

Then you just do the same thing as I described for #9. I understand you
may already know some of this, or most of it, or it may be hard to understand it
because it's in text, not a picture.

All due respect, I was just making sure you got the full explanation. I didn't do
that well of one but hope it helps!