Vertical or Horizontal ellipse?

[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

[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~

[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?

[Data]

indeed it is.

[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?
:smile:

[aisha]

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

[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?

[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 :wink:)

[aisha]

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

[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.

[aisha]

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

[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.