Question

Find the center and vertices of the hyperbola
16x^2 - 9y^2 - 96x + 72y - 144 = 0

Answer 1

your job is to make this look like
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]
do you know how to do that?

Answer 2

no I don't.. sorry i'm really lost!

Answer 3

ok we can go slow, but the first thing to know is that if you want the center and vertices you need to put it in the form i wrote above, that way you can read off the answer
if you make it look like \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] then the center is \((h,k)\)

Answer 4

this annoying task is done by "completing the square"
do you know how to do that? if the answer is "no" that is fine i will walk you through it

Answer 5

actually this one will not be so bad, because it was cooked up to give nice whole number answers
ready?

Answer 6

yeah I need help completing the square

Answer 7

ok so the first thing to do it get the constant \(-144\) on the right side of the equal sign, and also put the \(x\) terms and \(y\) terms next to each other

Answer 8

i.e. write
\[16x^2-96x-9y^2+72y=144\]

Answer 9

ok so far?

Answer 10

Yep sounds good

Answer 11

next step is to factor out the leading coefficient from the \(x\) terms and the \(y\) terms, which luckily is easy in this case

Answer 12

\[16(x^2-6x)-9(y^2-8y)=144\]

Answer 13

Oh okay i'm good so far

Answer 14

notice that when you factor out the \(9\) from the \(y\) terms it is \(-9(y^2-8y)\) by the distributive law, not for example \(-9(y^2+8y)\) a common mistake

Answer 15

now to complete the square
have of 6 is 3, and half of 8 is 4 so write
\[16(x-3)^2-9(y-4)^2=144+something\] we need to be careful about the something

Answer 16

but before we find the 'something" is the left hand side clear? i mean clear how to get it? it is not that hard in this case

Answer 17

Yeah I get it so far thank you

Answer 18

ok what comes next doesn't usually happen, so let walk through it in any case
we need to know what to add to the right, which is the same thing we added to the left when we completed the square

when you change \(x^2-6x\) in to \((x-3)^2\) you have added 9, because \((x-3)^2=x^2-6x+9\)
then when you multiply by 16 you see you have actually added \(9\times 16=144\)

you don't have to do all this thinking, you just need to think \(3^2\times 16=144\) so if you add 144 to the left, you have to add 144 to the right

Answer 19

then for the \(y\) term you have added \(4^2\times -9=-144\)

Answer 20

so in fact in this example when you completed the square you added NOTHING to the elft and right because \(144-144=0\) but this is very UNUSUAL so don't think it always works this way, almost always you have to add something to the right

Answer 21

now we are left with
\[16(x-3)^2-9(y-4)^2=144\] to set it equal to 1, divide both sides by 144 and write
\[\frac{(x-3)^2}{9}-\frac{(y-4)^2}{16}=1\]

Answer 22

almost done
you still following more or less?

Answer 23

Yeah it makes sense so far!

Answer 24

ok so we need the center, which is just a matter of reading\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]\[\frac{(x-3)^2}{9}-\frac{(y-4)^2}{16}=1\]so \(h=3,k=4\) and the center is \((3,4)\)

Answer 25

Okay I get that thanks, how do you get the vertices?

Answer 26

what else you need? foci or vertices?

Answer 27

oh ok vertices

Answer 28

for this we need to have some good idea what it looks like
since \(9<16\) it look like this (roughly)

Answer 29

you need to know it look like this to see that the vertices are to the left and right of the center, not up and down

Answer 30

in this example \(a^2=9\) and so \(a=3\)
therefore you go 3 units to the left and right of the center
the center is \((3,4)\) three units to the left is \((0,4)\) and 3 units to the right is \((6,4)\)

Answer 31

hope it is clear that all the information comes from looking at \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]

Answer 32

Yeah I understand that thank you so much

Answer 33

yw