Question

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7).

Answer 1

Is this hyperbola horizontal or vertical?

Answer 2

@Mertsj vertical

Answer 3

Good. That means the y term will be positive. Do you agree?

Answer 4

@Mertsj yes

Answer 5

And we know for a hyperbola that :
\[c^2=a^2+b^2\]

Answer 6

And we know that the distance from the center to the vertex is "a" and the distance from the center to the focus is "c"

Answer 7

We know that this hyperbola has center (0,0)

Answer 8

So its equation will look like this:
\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]

Answer 9

So now, you should be able to write its equation.

Answer 10

@Mertsj my only problem is I don't know what #'s I should plug in

Answer 11

What is the center of this hyperbola?

Answer 12

@Mertsj (0,0)?

Answer 13

Yes. Now you need to think about the vertex. What is the vertex?

Answer 14

@Mertsj I don't know

Answer 15

Allow me to suggest that you read the problem.

Answer 16

@Mertsj the vertex is (0,2)

Answer 17

How far is it from the center to the vertex?

Answer 18

@Mertsj 2 units

Answer 19

Do you remember that I said the distance from the center to the vertex is "a"?

Answer 20

@Mertsj so its 4

Answer 21

a^2=4. Yes.

Answer 22

if it is 4 then b = 49

Answer 23

No.

Answer 24

What is the focal point?

Answer 25

@Mertsj its 7

Answer 26

According to your problem, the focal points are (0,7) and (0,-7). That tells you that the focal points are 7 units from the center and that c =7. Use the relationship between a, b, and c to find b.

Answer 27

@Mertsj thanks for the enlightenment it will be 7^2=2^2+b^2 = 49=4+b : 49-4= 45

Answer 28

Perfect!! Now can you write the equation? What is it?

Answer 29

@Mertsj \[\frac{ y ^{2} }{ 4 }-\frac{ x ^{2} }{ 45 }=1\]

Answer 30

Wonderful!! You get an A for the day!!

Answer 31

@Mertsj thanks for the help

Answer 32

You're welcome. (I love conic sections)