Find the vertices and foci of the hyperbola with equation $\dfrac{(x-5)^2}{81} - \dfrac{(y-1)^2}{144} = 1$

Question

sleepyjess · Answer

I know the vertex is (5, 1)

sleepyjess · Answer

and the vertices are $(h, k \pm a)$ 

foci are $(h, k \pm c)$

sleepyjess · Answer

a is 9

c is 15

sleepyjess · Answer

So would the right answer be vertices: (5, 10) and (5, -8) foci: (5, 16) and (5, -14) ?

sleepyjess · Answer

@ganeshie8

amistre64 · Answer

vertex is not 5,1

amistre64 · Answer

center is 5,1

amistre64 · Answer

vertex is in line with the center, but its when they negative parts goes 0,  what solutions do we have for the waht remains?

amistre64 · Answer

your vertex formulas are not appropriate for the direction of this hyperbola ....

since we are subtracting the y parts, there is no vertex in the y direction

-y^2/144 = 1  is not possible

so our vertex has to be parallel to, long the direction of, the x axis

(x-5)^2/81 = 1

sleepyjess · Answer

I'm still confused :/ 

What's the difference between the center and vertex?

amistre64 · Answer

location ...

amistre64 · Answer

|dw:1430715817590:dw|

sleepyjess · Answer

Okay, so how do we find the vertex with the center? Or am I totally on the wrong track :/

sleepyjess · Answer

Or rather, vertices

amistre64 · Answer

well, we have to detemrine the way the hyperbola is directed, opening up in which directions,

sleepyjess · Answer

It's opening this way |dw:1430712424750:dw|