A very hard short ellipse question that came in my test today

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February 9th 2007, 01:27 AM
^_^Engineer_Adam^_^

A very hard short ellipse question that came in my test today

Given $F(2,3)$
$V(2,6)$
Minor Axis 6

How can i locate the center?
Is this possible? My classmates even didnt get this because they think it should be that the major axis will be given
:confused: :confused: :confused:
February 9th 2007, 02:49 AM
ticbol

Quote:

Originally Posted by ^_^Engineer_Adam^_^

Given $F(2,3)$
$V(2,6)$
Minor Axis 6

How can i locate the center?
Is this possible? My classmates even didnt get this because they think it should be that the major axis will be given
:confused: :confused: :confused:

If F signifies "focus", and V is for "vertex", then the vertical line x=2 is where the major axis is. Because a vertex is an endpoint of the major axis and a focus is in the major axis. The major axis can be described as vertex-focus-center-focus-vertex.

Okay, if that is all what is given, then the center can be (2,0).
"Can be" only, because as your classmates said, the major axis should have been given. As it is, the center can be anywhere from (2,2) down to (2,-infinity), if the locations are in integers.

So why (2,0)?
Because if the center were really at (2,0) and the minor axis is 6 units long, then the "dimensions" of the ellipses are multiples of 3. :)
Focal length (focus to near vertex) = 3 units.
Major axis = 12 units
Minor axis = 6 units
Lame reason, I know. :)
February 9th 2007, 05:08 AM
earboth

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Quote:

Originally Posted by ^_^Engineer_Adam^_^

Given $F(2,3)$
$V(2,6)$
Minor Axis 6
How can i locate the center?...

Hello,

let a be the length of the major axis
let b be the length of the minor axis
let e be the distance between centre and one focus

then you know:

1) b = 6
2) a - e = 3 that means a = e + 3
3) $e^2+b^2=a^2$

Plug in the values you know into 3):

$e^2+36=(e+3)^2=e^2+6e+9$

You'll get e = 4.5

Therefore the centre C has the coordinates: C(2, -1.5)

Therefore the equation of this ellipse is:

$\frac{(x-2)^2}{36}+\frac{(y+1.5)^2}{(7.5)^2}=1$

I've attached a diagram which shows the ellipse

EB
February 9th 2007, 11:22 PM
earboth

an additional remark

Quote:

Originally Posted by ^_^Engineer_Adam^_^

Given $F(2,3)$
$V(2,6)$
Minor Axis 6...

Hello,

As you have noticed certainly I have taken the length of the half minor axis as 6 units.
If you do the problem word-for-word then you must change my way to solve the problem in two points:

let a be the length of the half major axis
let 2b be the length of the minor axis
let e be the distance between centre and one focus

then you know:

1) 2b = 6 <==> b = 3
2) a - e = 3 that means a = e + 3
3) $e^2+b^2=a^2$

Plug in the values you know into 3):

$e^2+9=(e+3)^2=e^2+6e+9$

You'll get e = 0

That means there isn't any eccentricity. You get an "ellipse" with the two foci and the centre in one point: C = F_1 = F_2.
This "ellipse" is a circle with r = a = b = 3 and the centre C(2, 3).

EB