can anyone answer this?
find the center of this ellipse:
0=4x²+9y²−8x−18y−23
separate the values with commas.

- anonymous

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- anonymous

x^2/a^2 + y^2/b^2 = 1
This is long so I will do it in parts

- anonymous

Rearrange the terms and take the -23 to the other side so we can complete the square

- anonymous

23 = 4x^2 - 8x + 9y^2 - 18y

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## More answers

- anonymous

Factor out the 4 and the 9
23 = 4(x^2 - 2x ) + 9(y^2 - 2y )

- anonymous

Complete the square
23 = 4(x^2 - 2x + 1) + 9(y^2 - 2y + 1)

- anonymous

You actually added 4 and 9 to the right side because if you distribute this out you get 4(1) = 4 and 9(1) = 9
brb

- anonymous

OK... since you added 13 to the right side in order to complete the square, you need to add 13 to the left side
36 = 4(x^2 - 2x + 1) + 9(y^2 -2y + 1)

- anonymous

Now write the right side as ( )^2
36 = 4(x - 1)^2 + 9(y-1)^2

- anonymous

Because the formula is supposed to equal 1 you need to divide by 36
1 = 4(x - 1)^2/36 + 9(y - 1)^2/36

- anonymous

Cancel the 4 into the 36 and the 9 into the 36
1 = (x - 1)^2/9 + (y - 1)^2/9

- anonymous

Write the denominators as ( ) ^2
1 = (x - 1)^2/3^2 + (y - 1)^2/3^2
The center is (1,1)

- anonymous

on the center, do you change the signs?

- anonymous

Yes, that is how you find it.

- anonymous

Now, could you explain how to find the major axis vertices for this ellipse

- anonymous

Just a second

- anonymous

thanks!

- anonymous

In this case you actually have a circle because the denominators are the same.
Here is a good website for you to see.
http://www.mathwarehouse.com/ellipse/equation-of-ellipse.php
You use the ( )^2 on the bottom they are both 3^2
So you take the (3)^2 that is under the (x-1)^2 and you move 3 units to the right and to the left of the center.
You the take the (3)^2 that is under the (y-1)^2 and you move 3 units up and down from the center.

- anonymous

If you had a 5^2 on the bottom of the x^2 term and a 3^2 on the bottom of the y^2 term you would have major axis vertices - you would move 5 units right and left from the center and minor axis vertices 3 units up and down from the center.
Sorry. In you problem
Vertices at (4,1), (-2,1) and (1,4), (1,-2)

- anonymous

for the answer, it asks for the answer to be written (x1,y1)(x2,y2)

- anonymous

OOPS I made a mistake. Go back and look when I had a 9 by the (y-1)^2
The 9 goes into the 36 4 times so the bottom of that one should be (2)^2
So since the (x - 1)^2 has a 3^2 under it and the (y - 1)^2 has a (2)^2 under it
your x axis is your major axis

- anonymous

Equation is 1 = (x - 1)^2/3^2 + (y - 1)^2/2^2
Center (1,1)
Major axis vertices are (1 + 3,1) and (1-3,1) you move 3 units right and left from the center
(4,1) (-2,1)

- anonymous

using the same question. minor axis in the same answer form of (x1,y1)(x2,y2)

- anonymous

Minor would be the 2 units up and 2 units down from the center
(1, 1 + 2), (1, 1 - 2)
(1,3), (1,-1)

- anonymous

next is the foci

- anonymous

with the same answer form as the axis

- anonymous

Just a minute

- anonymous

OK you use the two numbers in the denominators, in this case the 3 and the 2
c^2 = a^2 + b^2 the denominators are the a and the b
c^2 = 3^2 + 2^2
c^2 = 9 + 4
c^2 = 13
c = sqrt(13)
So you move the sqrt(13) units from the center towards the major axis vertices.
So since the major axis is the x axis
(1 + sqrt13, 1), (1 - sqrt13, 1)

- anonymous

so the foci is (1 + sqrt13, 1), (1 - sqrt13, 1)

- anonymous

Yes

- anonymous

so it would be (4.61,1)(2.61,1)

- anonymous

(-2.61,1) and (4.61,1) you missed the negative sign

- anonymous

so just to make sure i understand this. can we both do the next few problems to make sure we get the same answer.

- anonymous

Sure

- anonymous

find the center of this hyperbola:
\[x ^{2}-y ^{2}-18x+10y+57=0\]
separate the values with commas.

- anonymous

This is interesting.. this is a hyperbola
x^2/a^2 - y^2/b^2 = 1
It can be
y^2/b^2 - x^2/a^2 = 1
You put the positive one first... In this problem the x^2 is positive

- anonymous

What do you get after you complete the square?
Be VERY careful on the y^2 one.

- anonymous

i'm kinda lost

- anonymous

x^2 - 18x - y^2 + 10y = -57
(x^2 - 18x ) - (y^2 - 10y ) = -57 do you understand this... I had to change the sign on the 10y term because there is a "-" between these squared terms and if you distributed the - sign back you need to get back to -y^2 + 10y
This is the hard step

- anonymous

(x^2 - 18x + 81) - (y^2 - 10y + 25) = -57 + 81 - 25
The reason it is -25 is because if you distributed it back out you actually added a "-25" to the left side.

- anonymous

i think why it's a little more difficult is because of the extra steps. it's confusing.

- anonymous

It is because of the "-" sign in the middle that makes it tricky.

- anonymous

(x - 9)^2 - (y - 5)^2 = -1 Are we ok to this point.

- anonymous

yeah

- anonymous

OK, the formula says it has to = 1 so we need to multiply every term by (-1) so it will = 1
-(x - 9)^2 + (y - 5)^2 = 1
Then the first term has to be positive with a minus between the two terms. so
(y - 5)^2 - (x - 9)^2 = 1 OK

- anonymous

so far so good

- anonymous

Center would (9,5) it is always (x,y)

- anonymous

do you change the signs? or is that already done?

- anonymous

It was (x - 9) so I changed it to 9
It was (x - 5) so I changed it to 5

- anonymous

next is to find the vertices for the hyperbola
& the answer in this (x1,y1),(x2,y2) form

- anonymous

Because the (Y-5)^2 is the positive term, the vertices are up and down from the center.
There is an implied (1)^2 under both the (Y-5)^2 and the (x - 9)^2 so you move 1 unit up and 1 unit down from the center.
(0,0+1), (0,0-1)
(0,1), (0,-1)

- anonymous

well i actually was doing a test. which i have been working on for almost a month because i only can miss 2 questions. which i already missed 3.

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