Question

Find the vertices and foci of the hyperbola with equation (x-3)^2/16 - (Y+4)^2/9=1
Vertices: (7, -4), (-1, -4); Foci: (-2, -4), (8, -4)
Vertices: (-4, 7), (-4, -1); Foci: (-4, -2), (-4, 8)
Vertices: (6, -4), (0, -4); Foci: (0, -4), (6, -4)
Vertices: (-4, 6), (-4, 0); Foci: (-4, 0), (-4, 6)

Answer 1

find the eccentricity first.

Answer 2

how?

Answer 3

so you have
$$
\cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\\
\cfrac{(x-3)^2a}{4^2}-\cfrac{(y+4)^2}{3^2}=1\\
$

can you tell which is your "a" element? that is, the one that will have the "traverse axis"

Answer 4

shoot

Answer 5

so you have
$$
\cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\\
\cfrac{(x-3)^2a}{4^2}-\cfrac{(y+4)^2}{3^2}=1\\
$$

can you tell which is your "a" element? that is, the one that will have the "traverse axis"

Answer 6

what the heck I have an "a" there? lemme repaste

Answer 7

so you have
$$
\cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\\
\cfrac{(x-3)^2}{4^2}-\cfrac{(y+4)^2}{3^2}=1\\
$$

can you tell which is your "a" element? that is, the one that will have the "traverse axis"

Answer 8

anyhow, the hyperbola is opening towards the axis with the POSITIVE sign

Answer 9

can you find the center of the hyperbola?

Answer 10

is pretty much all there GIVEN in the equation you have really

Answer 11

can you see them now?
$$
\cfrac{(x-\color{red}{h})^2}{\color{blue}{a}^2}-\cfrac{(y-\color{red}{k})^2}{b^2}=1\\
\cfrac{(x-\color{red}{(3)})^2}{\color{blue}{4}^2}-\cfrac{(y-\color{red}{(-4)})^2}{3^2}=1\\
$$

Answer 12

is this the answer? Vertices: (-4, 6), (-4, 0); Foci: (-4, 0), (-4, 6)

Answer 13

dunno

Answer 14

to find the vertices, you'd need to move "a" distance from the center of the hyperbola

Answer 15

over the traverse axis

Answer 16

ok

Answer 17

can you find the center in your equation? and the "a" element?