Question

find the vertices and foci for the hyperbola shown in the equation

Answer 1

here is the equation

Answer 2

and these are the choices

A) Vertices: (0, 3) and (3, 0); foci: (0, 0) and (0, 8)
B) Vertices: (-5, 0) and (5, 0); foci: (-8, 0) and (8, 0)
C) Vertices: (-3, 0) and (3, 0); foci: (-5, 0) and (5, 0)
D) Vertices: (-3, 0) and (3, 0); foci: (-8, 0) and (8, 0)

Answer 3

Is the attached enough for you to try to answer?

Answer 4

is there a difference between the equation of ellipse and that of a hyperbola ?

Answer 5

sorry but I'm a bit confused with the attached, can you narrow it down please

Answer 6

The way you tell the difference is that when in standard form, the hyperbola has one of the terms negative. That is obviously the "y" term in your problem.

Answer 7

Since you have the x-term positive, the so-called transverse axis is horizontal. that means that the center, foci, and vertices are all on a horizontal one-- the x-axis in your problem.

Answer 8

ok .....

Answer 9

The center is at (0, 0) and the vertices are a units aways from the center, in general (-a, 0) and (a, 0). In your problem a = 3 (a^2=9).

Answer 10

So your vertices are (?, 0) and (?, 0)

Answer 11

Your turn :)

Answer 12

yes so I have the (3,0_ and (-3,0) as the vertices , right ?

Answer 13

Right!

Answer 14

sorry (3,0) (-3,0)

Answer 15

good :)

Answer 16

now the foci

Answer 17

Now the 55 is b^2 (refer back to the graphic I posted earlier. We will need that to fond the foci. Again, in that graphic, the foci (focus is singular) are calculated using the a and b.

Answer 18

ok ...

Answer 19

Textbooks usually use the letter c to indicate the distance the foci are from the center. Btw, it is not shown on the graphic above. The a, b, and c are in a Pythagorean relationship such that c^2 = a^2 + b^2 for any hyperbola. Can You use that to find c? The foci are (-c, 0) and (c, 0).

Answer 20

yes so it will be c^2= a^2 + b^2
so c^2= 9+55= 64
so c= 8
then the foci are (-8,0) and (8,0)

Answer 21

That's it!

Answer 22

that was really really helpful , thank you so much for your help @gryphon

Answer 23

You are welcome. One last thing. If the negative sign were in front of the x-term, then the orientation of the hyperbola would be "vertical." That is instead of looking like parentheses opening the wrong way, it would look like a smiley and sad face smile.

Answer 24

got it, if it is "vertical" instead. I 've already saved the graph. thank you

Answer 25

OK, see ya later!

Answer 26

see ya , thanks again :)