Question

identify the conic 5x^2+9y^2-40x-18y+44=0 and give the coordinates of its center

Answer 1

so, the \(x^2 \) and \(y^2\) components are POSITIVE, that's a dead giveaway is an ellipse

Answer 2

well, could be a circle, or an same-axis values ellipse ehehe

Answer 3

the rest is just "completing the square"

Answer 4

do you know how to "complete the square" to get a "perfect square trinomial"?

Answer 5

I know how to complete the square normally...do you do something special to get a perfect square trinomial?

Answer 6

Yep, do that 3 times

Answer 7

well, you grab all the "x" elements and the "y" elements and group them together first
get a common factor so you're only left inside the parentheses with \(x^2 \ and\ y^2\)
as opposed to \(3x^2 \ and \ 5y^2\)

Answer 8

from there, we work with our good jolly friend, Mr 0 [zero]

Answer 9

ugh I hate completing the square! ok give me a minute

Answer 10

I'll rewrite it some in a few secs

Answer 11

$$
5x^2+9y^2-40x-18y+44=0\\
(5x^2-40x)+(9y^2-18y)+44\\
\text{so get common factors}\\
5(x^2-8x+\boxed{?})+9(y^2-2y+\boxed{?})+44=0\\
$$

Answer 12

so, we need to find those " ? " values, and those numbers come from Mr 0
so, if you find 25 and 13, you'd need to SUBSTRACT 25 and 13
because what we're really doing is BORROWING from 0
0 = 25+13-25-13 :)

Answer 13

wait wait wait. what are 25 and 13?

Answer 14

where did they come from?

Answer 15

just an example

Answer 16

oh ok. I was totally lost

Answer 17

just a couple of arbitrary numbers ehhe

Answer 18

so i need to plug in 0 where the ?'s are?

Answer 19

one sec

Answer 20

$$
5x^2+9y^2-40x-18y+44=0\\
(5x^2-40x)+(9y^2-18y)+44\\
\text{so get common factors}\\
5(x^2-8x+\color{blue}{4^2})+9(y^2-2y+\color{blue}{1^2})+44=0\\
\text{so plus } 5(4^2)+9(1^2), \ \ then \ \
\color{red}{-5(4^2)-9(1^2)}\\
5(x^2-8x+\color{blue}{4^2})+9(y^2-2y+\color{blue}{1^2})
\color{red}{-5(4^2)-9(1^2)}+44 = 0
$$

Answer 21

so, we're really borrowing from 0, +89, and giving the loan back with -89

Answer 22

ok what next?

Answer 23

the parenthesized "perfect squares" are just a 2 binomials, so hold on

Answer 24

$$
5(x-4)^2+9(y-1)^2
\color{red}{-89}+44 = 0\\
5(x-4)^2+9(y-1)^2 = 45
$$

Answer 25

from there you just make the = 45, to =1
or set the equation to = 1
meaning divide both sides by 45
that is each equation over 45 = 45/45

Answer 26

ok
and that's it?

Answer 27

$$
\cfrac{5(x-4)^2}{45}+
\cfrac{9(y-1)^2}{45}=1\\
\cfrac{(x-4)^2}{9}+
\cfrac{(y-1)^2}{5}=1\\
$$

Answer 28

and that's your ellipse

Answer 29

ok thank you!
you are so helpful

Answer 30

and the coordinates for the center is at (h,k)

http://www.mathwarehouse.com/ellipse/images/translations/general_formula_major.gif

Answer 31

yw

Answer 32

so center is (-4,-1)

Answer 33

well, (x- (+4)) => (x-4)
and (y - (+1) => (y-1)

Answer 34

so, the " - " isn't part of the h,k elements

Answer 35

so just (4,1)

Answer 36

is part of the formula
now if you had (x+4) => (x-(-4)) so the "h" woulld be -4 then

Answer 37

yes, (4,1)