# Homework Help: HELP! Features of A Hyperbola CONIC!

1. Apr 2, 2005

### aisha

$$\frac {(x-1)^2} {9} - \frac {(y+2)^2} {25} = 1$$

I think this is a vertical hyperbola
with center (1,-2) a=5 b=3

Transversal length=10
Conjugate length =6

Vertex (1,-7) and (1,3)
Foci (1,sqrt(34)-2) and (1,-sqrt(34)-2)
Asymptotes y+2=5/3(x-1) and y+2=-5/3 (x-1)

I dont know if all these values are correct I need to find the intercepts if there are any can someone show me how?

2. Apr 2, 2005

### Jameson

If you need to find the intercepts set x and y equal to zero and solve for the other point in each case. That's if you want to find the points (0,y) and (x,0)

EDIT: Just to clarify, I meant you need set each variable equal to zero to find the intercepts separately, not at once.

$$\frac {(0-1)^2} {9} - \frac {(y+2)^2} {25} = 1$$

and

$$\frac {(x-1)^2} {9} - \frac {(0+2)^2} {25} = 1$$

Last edited: Apr 2, 2005
3. Apr 3, 2005

### aisha

When I set y=0 I got x=-7 or x=7
But when I set x=0 and solved for y I didnt get something as simple as x maybe I did something wrong, I think there are no y-intercepts am I correct?
When I solve for y with x=0 I get an imaginary number

4. Apr 3, 2005

### Jameson

Sketch the graph and think analytically about if there will be a x or y intercept. You have all of the information needed to draw the graph.

5. Apr 3, 2005

### aisha

Are all of my features correct? can someone please check!

6. Apr 3, 2005

### Jameson

The ones I checked looked ok, I didn't look at them all though. Your center, transverse length, and conjugate length are correct. Show me where you think you messed up.

7. Apr 3, 2005

### aisha

for the yintercept I got

y=+- Sqrt(-200) -2

when I set x=0 and solved for y. Are my x-intercepts correct? Can you check my foci vertices and asymptotes too.

8. Apr 3, 2005

### Hurkyl

Staff Emeritus
Nitpick: instead of saying "a=5, b=3", you should state to what those variables actually refer.

This particular example is particularly important, because a lot of people (meaning me!) will immediately think of the usual form: $(x-h)^2/a^2 - (y-k)^2/b^2 = 1$.

9. Apr 3, 2005

### aisha

I think ive made a huge mistake all of the features should be for the horizontal hyperbola not vertical hyperbola, right?

10. Apr 3, 2005

### Data

indeed, it is a horizontal hyperbola (if by horizontal, you mean opens sideways).

11. Apr 3, 2005

### aisha

I have changed all my features but still having problems solving for the x-intercepts plz help me what do i do with the denominators?

$$\frac {(x-1)^2} {9} - 4/25=1$$

$$\frac {(x-1)^2} {9}=21/25$$

I dont know what to do with the 9 what is the final answer?

12. Apr 3, 2005

### Jameson

Multiply by 9 and you get:

$$(x-1)^2 = \frac{21}{25}*9$$
$$x - 1 = \sqrt {\frac{21}{25}*9}$$
$$x = \sqrt {\frac{21}{25}*9} + 1$$

Is this what you were having trouble with?

13. Apr 3, 2005

### Data

mulpliply both sides by 9... bu what you actually mean is

$$\frac{(x-1)^2}{9} - \frac{4}{25} = 1 \Longrightarrow \frac{(x-1)^2}{9} = \frac{29}{25}$$

so you can then just say

$$(x-1)^2 = 9\frac{29}{25} \Longrightarrow x-1 = \pm\frac{3\sqrt{29}}{5} \Longrightarrow x = 1 \pm \frac{3\sqrt{29}}{5}$$