Question

Find the vertices and foci of the hyperbola with equation ((x-5)^2/81)-((y-1)^2)/144)=1??? help please :c

Answer 1

\[\frac{(x-5)^2}{9^2}-\frac{(y-1)^2}{12^2}=1\] is a start
the center is therefore \((5,1)\)

Answer 2

you need to know the center in order to find the vertices
you also need to know what it looks like
do you know what this one looks like?

Answer 3

no :/

Answer 4

because the \(x^2\) term comes first, it looks like this

Answer 5

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

Answer 6

general form is
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]and you have \[\frac{(x-5)^2}{9^2}-\frac{(y-1)^2}{12^2}=1\] so \(a=9\)

Answer 7

this tells you the vertices are 9 units to the left and right of the center, which is \((5,1)\)

Answer 8

im stilll confused :/ im sorry

Answer 9

do you see tjat \(a^2=81\) so \(a=9\) ?

Answer 10

yeah

Answer 11

ok and do you see from the equation that the center is \((5,1)\) ?

Answer 12

yes, but for the vertices would i just add and subtract 9 from 5?

Answer 13

yes

Answer 14

that is why you need the center first

Answer 15

what else do you need? the foci?

Answer 16

yeah the foci, and just to double check would the vertices be (14,1) and (-4,1)?

Answer 17

yes

Answer 18

for the foci, again they will be to the left and right of the center, but you need another number

Answer 19

\[c^2=a^2+b^2\] just like pythagoras
then in your case
\[c^2=81+144=225\] and so \(c=15\)
now go 15 units left and right for the foci

Answer 20

okay, thanks so much!(:

Answer 21

yw

Answer 22

btw if you ever get stuck, there are real good worked out examples here
http://www.purplemath.com/modules/hyperbola2.htm

Answer 23

and if you need to cheat, you can check your answers here
http://www.wolframalpha.com/input/?i=hyperbola++%28%28x-5%29^2%2F81%29-%28%28y-1%29^2%29%2F144%29%3D1