Question

Find the graphing form, coordinates of the center & foci, the length of the major and minor axis, then graph the points and equation. 16x2 – 32x + 25y2 + 100y – 284 = 0

Answer 1

@KendrickLamar2014

Answer 2

\[16x^2 – 32x + 25y^2 + 100y – 284 = 0\]is going to take a while
how much time you got?

Answer 3

the whole night

Answer 4

that is to say, we can do it in one step, or grind it out to make it look like
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 5

you are no doubt supposed to do it the long algebra way, which requires factoring and then completing the square twice

Answer 6

right

Answer 7

\[16x^2 – 32x + 25y^2 + 100y – 284 = 0\\
16x^2 – 32x + 25y^2 + 100y =284\\
16(x^2 – 2x) + 25(y^2 + 4y) = 284\] is a start

Answer 8

you good so far?

Answer 9

this is where you say "yeah got that"

Answer 10

lol, i was reviewing it, yeah, got that

Answer 11

ok so now to complete the square
half of 2 is 1, half of 4 is 2, first we go right to
\[16(x-1)^2+25(y+2)^2=284+?+?\]

Answer 12

just gotta figure out what we added to the left so we can add the same thing to the right
hence the two question marks

Answer 13

2?

Answer 14

heck no

Answer 15

\[(x-1)^2=x^2-2x+1\] multiply 1 by 16 we added 16 just by changing
\[16(x^2-2x)\] in to
\[16(x-1)^2\]

Answer 16

similarly
\[(y+21)^2=y^2+4y+4\] multiply 4 by 25 we added 100 by changing
\[25(y^2+4y)\] in to
\[25(y+2)^2\]

Answer 17

typo there, but you get the idea
that 21 should be a 2

Answer 18

okay, one second let me look through this

Answer 19

\[16(x-1)^2+25(y+2)^2=284+?+?\\
16(x-1)^2+25(y+2)^2=284+16+100\]
take your time

Answer 20

let me know when you want to continue, there is really only one more step after adding

Answer 21

okay lets continue

Answer 22

add

Answer 23

\[16(x-1)^2+25(y+2)^2=284+16+100\\
16(x-1)^2+25(y+2)^2=400\]

Answer 24

then divide both sides by \(400\) to make the right hand side a 1, and it will be in standard form

Answer 25

you got that, or you want me to write it?

Answer 26

both sides by 400? yeah write it

Answer 27

the first step is
\[\frac{16(x-1)^2}{400}+\frac{25(y+2)^2}{400}=\frac{400}{400}\]but the point is you get
\[\frac{(x-1)^2}{25}+\frac{(y+2)^2}{16}=1\]

Answer 28

all that work to make it look like \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] but now everything should be easy

Answer 29

you see from your eyeball that \(h=1,k=-2\) so you know the center is \((1,-2)\)

Answer 30

i forget what else you need

Answer 31

foci and the length of the major and minor axis

Answer 32

any idea on how to make graphs on libre?

Answer 33

ok first question, does it look like the one on the left or on the right?

Answer 34

"i have no idea' is a fine answer, i can tell you and tell you how you know it

Answer 35

the second one?

Answer 36

nope

Answer 37

bigger number is under the \(x\) term, so the first one

Answer 38

we need to know this to find the vertices
in this case \(a^2=24\) so \(a=5\) since it looks like the one in the left, that means the vertices are 5 units to the LEFT AND RIGHT of the center

Answer 39

if it was oriented the other way, then the vertices would be up and down, not left and right

Answer 40

okay im following you so far

Answer 41

if i lose you let me know
but you see how important it is to know which way it is oriented, so you don't get all screwed up with left/right vs up/down

Answer 42

now to find the foci, we need another number,
we have
\[a^2=25,b^2=16\] we need
\[c^2=\sqrt{a^2-b^2}=\sqrt{25-16}=\sqrt9=3\] i.e just like pythagoras, the famous 3 - 4 - 5 right triangle

Answer 43

that means the foci are 3 units to the left and right of the center

Answer 44

3 units to the left of \((1,-2)\) is \((-2,-2)\) and 3 units to the right is \((4,-2)\)

Answer 45

hm, one second

Answer 46

i'll hold
if you have a question ask, or review

Answer 47

okay im good now

Answer 48

ok all that is left is the lenght of the major and minor axis, but we already have the major one,5 units to the left and right of the center for a total length of 10
and since \(b^2=16\) we have \(b=4\) so 4 units up and down from the center for a total length of 8