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

- anonymous

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.