Given $\displaystyle F(2,3)$

$\displaystyle V(2,6)$

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:

- Feb 9th 2007, 01:27 AM^_^Engineer_Adam^_^A very hard short ellipse question that came in my test today
Given $\displaystyle F(2,3)$

$\displaystyle V(2,6)$

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: - Feb 9th 2007, 02:49 AMticbol
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. :) - Feb 9th 2007, 05:08 AMearboth
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) $\displaystyle e^2+b^2=a^2$

Plug in the values you know into 3):

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

You'll get e = 4.5

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

Therefore the equation of this ellipse is:

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

I've attached a diagram which shows the ellipse

EB - Feb 9th 2007, 11:22 PMearbothan additional remark
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) $\displaystyle e^2+b^2=a^2$

Plug in the values you know into 3):

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

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