Question

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

Answer 1

x^2/a^2 + y^2/b^2 = 1

This is long so I will do it in parts

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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)

Answer 8

Now write the right side as ( )^2

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

Answer 9

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

Answer 10

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

Answer 11

Write the denominators as ( ) ^2

1 = (x - 1)^2/3^2 + (y - 1)^2/3^2

The center is (1,1)

Answer 12

on the center, do you change the signs?

Answer 13

Yes, that is how you find it.

Answer 14

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

Answer 15

Just a second

Answer 16

thanks!

Answer 17

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.

Answer 18

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)

Answer 19

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

Answer 20

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

Answer 21

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)

Answer 22

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

Answer 23

Minor would be the 2 units up and 2 units down from the center

(1, 1 + 2), (1, 1 - 2)

(1,3), (1,-1)

Answer 24

next is the foci

Answer 25

with the same answer form as the axis

Answer 26

Just a minute

Answer 27

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)

Answer 28

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

Answer 29

Yes

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Sure

Answer 34

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

Answer 35

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

Answer 36

What do you get after you complete the square?

Be VERY careful on the y^2 one.

Answer 37

i'm kinda lost

Answer 38

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

Answer 39

(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.

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

yeah

Answer 44

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

Answer 45

so far so good

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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)

Answer 51

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.