Question

Find the vertices and foci of the hyperbola with equation:
(x+1)^2/16 - (y+5)^2/9 = 1

Answer 1

is there a solution when x part = 0?

is there a solution when y part = 0?

Answer 2

might be better stated as:

(x+1) = 0
(y+5) = 0

Answer 3

Okay I see...
It just really wants me to write
Vertices: ( x, y) and (x, y) then foci: (x,y) and (x, y)
@amistre64

Answer 4

yes, but they have gone and moved this off the origin ....

assume the x parts are 0\[-\frac{[~y~]^2}{9}=1\]

has no "real" solution so the vertex will have to be when the y parts are zero; or when y = -5

Answer 5

by zeriong out the y parts we are left with\[\frac{(x+1)^2}{16}=1\]
what solutions do we have for this?

Answer 6

the foci will be relatively simple to determine

Answer 7

I feel like I'm kind of wasting your time, Amistre... Idk how to find the solutions for that there :(

Answer 8

at the moment we have

Vertices: ( x,-5) and (x, -5)

foci: (x,-5) and (x, -5)

Answer 9

would you agree that anything divided by itself is equal to 1? except for 0/0 of course

Answer 10

yes. and anything divided by the negative version or negative by positive version is -1

Answer 11

\[\frac{(x+1)^2}{16}=1\]
\[\frac{(x+1)^2}{16}=\frac{16}{16}\]
\[(x+1)^2=16\]

Answer 12

what squared, equals 16?

Answer 13

4

Answer 14

4 is one solution, but isnt (-4)^2 = 16 true as well?

Answer 15

yes so +- 4

Answer 16

good, then we know that

x+1 = 4

or

x+1 = -4

what solutions do we have for x now?

Answer 17

x = 3 or -5

Answer 18

good, then we know the vertexes

verts: (-5,-5) and (3,-5)

just have to determine the foci

Answer 19

okay lets do this

Answer 20

add the under parts together .... and sqrt the results, we will need this for a measurement

Answer 21

the general hyperB eq is
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

the foci have a measurement from the center that is equal to \(\sqrt{a^2+b^2}\)

Answer 22

okay but we never wrote x^2 and y^2 over anything