# Vertical or Horizontal ellipse?

1. Apr 2, 2005

### aisha

$$16x^2+9y^2+192y-36y+468=0$$

Was the original conic i had to conver this into standard form and got

$$\frac {(x+6)^2} {9} + \frac {(y-2)^2} {16} =1$$

Im not sure if this is a horizontal or vertical ellipse

2. Apr 2, 2005

### Data

the center of the ellipse is $(-6, 2)$. If you get $x=-6$, then what range of values can $y$ take? If you set $y = 2$ then what range of values can $x$ take?

Can these facts help you to decide? If you can't see why they can directly, then draw a picture of the ellipse and see if you can tell~

3. Apr 2, 2005

### aisha

since the denominator of the x is less that the denominator of the y then the equation is in the form

$$\frac {(x-h)^2} {b^2} + \frac {(y-h)^2} {a^2}$$

I think if what I said is right then this is a vertical ellipse

a=4 b=3?

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

### Data

indeed it is.

5. Apr 2, 2005

### aisha

Here are the features i got for this conic
Vertical ellipse
Center (-6,2) a=4 b=3
Length of major axis 2a=8
Length of minor axis 2b=6
Vertices (-6,-2) and (-6,6)
Foci=(-6,-3) and (-6,7) where c=5

are all of these correct?

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

### aisha

The focus is outsie of the ellipse? What is wrong? Is this ok?

7. Apr 3, 2005

### aisha

(h,-c+k) and (h,c+k) are the foci because this is a vertical ellipse

c=5 and I plugged in the center (-6,2) wats wrong?

8. Apr 3, 2005

### Data

what's $c$, and why do you think it's $5$?

If you're using it that way, then it should be $\sqrt{7}$ (I made a mistake earlier, by the way... that's why the other post is gone now )

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

### aisha

c= sqrt(a^2+b^2) how is it sqrt7?

10. Apr 3, 2005

### Data

it's actually

$$b^2 = a^2 - c^2 \Longrightarrow c^2 = a^2 - b^2 \Longrightarrow c = \sqrt{a^2-b^2}$$

using your formula, you would always find the focii outside the ellipse.

11. Apr 4, 2005

### aisha

Thanks soo much I made the same mistake in 3 problems now I rememeber thankssssssss LIFE SAVER!!

12. Apr 4, 2005

### vitaly

If the value under y^2 is greater, then it's going to be a vertical ellipse.
If the value under x^2 is greater, then it's going to be horizontal.

You can check by graphing and calculating the lengths of the major and minor axes. That should help you too.