Question

Find the vertices and foci of the hyperbola with equation

Answer 1

@electrokid Can you help?

Answer 2

@hartnn

Answer 3

1) what type of an algebraic curve does it look like?

Answer 4

oh, its give, a hyperbola

Answer 5

Yes a hyperbola

Answer 6

that canonical form gives you (h,k)
and a, b

Answer 7

Yes, but how do I know whether the hyperbola is a vertical or horizontal one?

Answer 8

is a^2 bigger or b^2

Answer 9

and what is "-ve" x^2 or y^2

Answer 10

a^2

Answer 11

since x is +ve and y is subtracted, we have x^2-y^2=1

Answer 12

so, the axis of the hyperbola is "x" axis

Answer 13

so, this is

Answer 14

not numbers there :)

Answer 15

but you got the idea?

Answer 16

Oh ok

Answer 17

So it's horizontal :)

Answer 18

yep

Answer 19

if y^2 was positive, "y" axis would be the axis of the hyperbola

Answer 20

Can you help me find the final answer, I keep trying but I'm getting answers that are not in any of the answer choices? @electrokid

Answer 21

@electrokid

Answer 22

(h,k)=(3,-4)
\(c = \sqrt{a^2+b^2}=\sqrt{9+16}=5\)
foci = \((h\pm c,k)=(3\pm5,-4)
=(-2,-4) and (8,-4)\)

Answer 23

seems like there is only one such option

Answer 24

What?

Answer 25

did you get the foci?

Answer 26

No

Answer 27

Do you still need help with this problem?

Answer 28

Yes, please!

Answer 29

Should we just start over or what?

Answer 30

If we can start over, with a detailed explanation, that would be really great. Thanks :)

Answer 31

You can get a lot of information about a hyperbola from its equation.

Answer 32

Yes

Answer 33

h, k, a, b

Answer 34

The center is (h,k)

Answer 35

The number below the positive term is a^2

Answer 36

The number below the negative term is b^2

Answer 37

So in this case, is (h,k)= (3, -4) and is a^2= 9 and b^2= 16 ?

Answer 38

For a hyperbola, the relationship between a, b, and c is a^2+b^2=c^2

Answer 39

Yes

Answer 40

You're just about right. You have the center right but it is the x term that is positive and the y term is negative so a^2=16 and b^2=9

Answer 41

The reason that a is so important is that the vertices are a units to the right and a units to the left of the center.

Answer 42

Are you with me?

Answer 43

Yes

Answer 44

So if a^2 = 16, then a = ?

Answer 45

So once I have the center, a (which is 4), b (which is 3), and c (which I believe is 5?), how do I use these to calculate the foci and vertices?

Answer 46

So yes, a is 4. The vertices are "a" units to the right and "a" units to the left of the center.

Answer 47

The center is (3,-4) . What point is 4 units to the right of (3,-4) ?

Answer 48

(7, -4)

Answer 49

oh

Answer 50

and (-1, -4) to the left

Answer 51

Very good. So now you know the vertices.

Answer 52

Yes, foci next

Answer 53

Now...the focal points. You correctly found that c is 5. That is important because the focal points are "c" units to the right and "c" units to the left of the center. So can you find the focal points?

Answer 54

So... (8, -4) and (-2, -4) ?

Answer 55

Very good.