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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2/a^2 + y^2/b^2 = 1 This is long so I will do it in parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Rearrange the terms and take the 23 to the other side so we can complete the square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.023 = 4x^2  8x + 9y^2  18y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Factor out the 4 and the 9 23 = 4(x^2  2x ) + 9(y^2  2y )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Complete the square 23 = 4(x^2  2x + 1) + 9(y^2  2y + 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0OK... 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
 5 years ago
Best ResponseYou've already chosen the best response.0Now write the right side as ( )^2 36 = 4(x  1)^2 + 9(y1)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because 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
 5 years ago
Best ResponseYou've already chosen the best response.0Cancel the 4 into the 36 and the 9 into the 36 1 = (x  1)^2/9 + (y  1)^2/9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Write the denominators as ( ) ^2 1 = (x  1)^2/3^2 + (y  1)^2/3^2 The center is (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on the center, do you change the signs?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that is how you find it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, could you explain how to find the major axis vertices for this ellipse

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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/equationofellipse.php You use the ( )^2 on the bottom they are both 3^2 So you take the (3)^2 that is under the (x1)^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 (y1)^2 and you move 3 units up and down from the center.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0for the answer, it asks for the answer to be written (x1,y1)(x2,y2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OOPS I made a mistake. Go back and look when I had a 9 by the (y1)^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
 5 years ago
Best ResponseYou've already chosen the best response.0Equation is 1 = (x  1)^2/3^2 + (y  1)^2/2^2 Center (1,1) Major axis vertices are (1 + 3,1) and (13,1) you move 3 units right and left from the center (4,1) (2,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using the same question. minor axis in the same answer form of (x1,y1)(x2,y2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Minor would be the 2 units up and 2 units down from the center (1, 1 + 2), (1, 1  2) (1,3), (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with the same answer form as the axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the foci is (1 + sqrt13, 1), (1  sqrt13, 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it would be (4.61,1)(2.61,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(2.61,1) and (4.61,1) you missed the negative sign

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so just to make sure i understand this. can we both do the next few problems to make sure we get the same answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the center of this hyperbola: \[x ^{2}y ^{2}18x+10y+57=0\] separate the values with commas.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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
 5 years ago
Best ResponseYou've already chosen the best response.0What do you get after you complete the square? Be VERY careful on the y^2 one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^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
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0i think why it's a little more difficult is because of the extra steps. it's confusing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is because of the "" sign in the middle that makes it tricky.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x  9)^2  (y  5)^2 = 1 Are we ok to this point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Center would (9,5) it is always (x,y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you change the signs? or is that already done?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It was (x  9) so I changed it to 9 It was (x  5) so I changed it to 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0next is to find the vertices for the hyperbola & the answer in this (x1,y1),(x2,y2) form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because the (Y5)^2 is the positive term, the vertices are up and down from the center. There is an implied (1)^2 under both the (Y5)^2 and the (x  9)^2 so you move 1 unit up and 1 unit down from the center. (0,0+1), (0,01) (0,1), (0,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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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