anonymous
  • anonymous
can anyone answer this? find the center of this ellipse: 0=4x²+9y²−8x−18y−23 separate the values with commas.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
x^2/a^2 + y^2/b^2 = 1 This is long so I will do it in parts
anonymous
  • anonymous
Rearrange the terms and take the -23 to the other side so we can complete the square
anonymous
  • anonymous
23 = 4x^2 - 8x + 9y^2 - 18y

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anonymous
  • anonymous
Factor out the 4 and the 9 23 = 4(x^2 - 2x ) + 9(y^2 - 2y )
anonymous
  • anonymous
Complete the square 23 = 4(x^2 - 2x + 1) + 9(y^2 - 2y + 1)
anonymous
  • 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
  • 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
  • anonymous
Now write the right side as ( )^2 36 = 4(x - 1)^2 + 9(y-1)^2
anonymous
  • 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
  • anonymous
Cancel the 4 into the 36 and the 9 into the 36 1 = (x - 1)^2/9 + (y - 1)^2/9
anonymous
  • anonymous
Write the denominators as ( ) ^2 1 = (x - 1)^2/3^2 + (y - 1)^2/3^2 The center is (1,1)
anonymous
  • anonymous
on the center, do you change the signs?
anonymous
  • anonymous
Yes, that is how you find it.
anonymous
  • anonymous
Now, could you explain how to find the major axis vertices for this ellipse
anonymous
  • anonymous
Just a second
anonymous
  • anonymous
thanks!
anonymous
  • 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
  • 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
  • anonymous
for the answer, it asks for the answer to be written (x1,y1)(x2,y2)
anonymous
  • 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
  • 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
  • anonymous
using the same question. minor axis in the same answer form of (x1,y1)(x2,y2)
anonymous
  • 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
  • anonymous
next is the foci
anonymous
  • anonymous
with the same answer form as the axis
anonymous
  • anonymous
Just a minute
anonymous
  • 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
  • anonymous
so the foci is (1 + sqrt13, 1), (1 - sqrt13, 1)
anonymous
  • anonymous
Yes
anonymous
  • anonymous
so it would be (4.61,1)(2.61,1)
anonymous
  • anonymous
(-2.61,1) and (4.61,1) you missed the negative sign
anonymous
  • 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
  • anonymous
Sure
anonymous
  • anonymous
find the center of this hyperbola: \[x ^{2}-y ^{2}-18x+10y+57=0\] separate the values with commas.
anonymous
  • 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
  • anonymous
What do you get after you complete the square? Be VERY careful on the y^2 one.
anonymous
  • anonymous
i'm kinda lost
anonymous
  • 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
  • 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
  • anonymous
i think why it's a little more difficult is because of the extra steps. it's confusing.
anonymous
  • anonymous
It is because of the "-" sign in the middle that makes it tricky.
anonymous
  • anonymous
(x - 9)^2 - (y - 5)^2 = -1 Are we ok to this point.
anonymous
  • anonymous
yeah
anonymous
  • 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
  • anonymous
so far so good
anonymous
  • anonymous
Center would (9,5) it is always (x,y)
anonymous
  • anonymous
do you change the signs? or is that already done?
anonymous
  • anonymous
It was (x - 9) so I changed it to 9 It was (x - 5) so I changed it to 5
anonymous
  • anonymous
next is to find the vertices for the hyperbola & the answer in this (x1,y1),(x2,y2) form
anonymous
  • 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
  • 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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