How can i know that this equation is a ellipse?

  • Thread starter Fabio010
  • Start date
  • Tags
    Ellipse
In summary, the problem is to sketch a region that is represented by the equation |z+i|+|z-1|=2 in the complex plane. This equation results in a simplified form of 3x^2+3y^2 = 4x-4y+2xy, which can be further simplified by rotating the coordinate system and completing the squares. The resulting equation is an ellipse with foci at -i and 1, indicating that the total distance from any point on the ellipse to these two foci is a constant of 2.
  • #1
Fabio010
85
0
Sketch the following region:

| z+i | + | z-1 | =2




I know that it is a ellipse, but i am trying to replace the z for x +iy.



I reached the following equation.

3x^2+3y^2 = 4x-4y+2xy


now how can a simplify it in order to have the ellipse equation: (x^2/a^2) + (y^2/b^2) = 1
 
Mathematics news on Phys.org
  • #2
Supposing you have done this correctly:

1. There is no reason that the axes of the ellipse should lie parallell to the x and y-axes.
That this is NOT the case, is given by the mixed quadratic term 2xy.
You need to rotate your coordinate system by some fixed angle "a", by introducing new, general coordinates (u,v), for example:
x=u*cos(a)-v*sin(a), y=u*sin(a)+v*cos(a)
We then determine the angle "a" so that the cross term in "u" and "v" vanishes.
Here, "a" is the angle between the positive x-axis and the positive u-axis, so that if "a"=0 yields x=u, y=v, while "a"=pi/2 in radians, x=-v and y=u.

2. Furthermore: The origin of the ellipse will NOT lie at the origin. Complete the squares in u and v to get (after a while) an equation of the form:
(u-u')^2/K^2+(v-v')^2/L^2=1

where u', v', K and L are constants to be determined.
 
Last edited:
  • #3
Now, following 1, you'll get:
3(u^2+v^2)=4(u(cos(a)-sin(a))-v(cos(a)+sin(a)))+2((u^2-v^2)cos(a)sin(a)+uv(cos(2a)).

Choosing a=pi/4, then, yields:
3(u^2+v^2)=-4sqrt(2)*v+u^2-v^2, that is:
2u^2+4(v^2+sqrt(2)v)=0

Now, you need to complete the square in v, according to 2.

This gives you:
2u^2+4(v+1/sqrt(2))^2=2,

That is:
u^2/K^2+(v+1/sqrt(2))^2/L^2=1, (**)
where u'=0, v'=-1/sqrt(2), K=1, L=1/sqrt(2)

(**) is your equation for the ellipse, where the u-axis makes pi/4 radians angle to the x-axis

The origin of the ellipse is therefore at (u',v')=(0,-1/sqrt(2)), which works out to be at (x',y')=(1/2,-1/2)

In order to draw it, you ought to find the extremals along the (u,v) axis pairs, and then work out the (x,y) coordinate pairs.
For example,
(u,v)=(0,0) works out as..(x,y)=(0,0)
 
Last edited:
  • #4
The geometric definition of 'ellipse' is "a plane figure having two points, called foci, such that the total distances from any point on the ellipse to the two foci is a constant".

In the complex plane, geometrically, |a- b| is the distance between point a and point b.

so |z+ i|+ |z- 1|= 2 says exactly that the total distance from any point, z, on the figure to -i and 1, is a constant, 2. That tells us it is an ellipse, having -i and 1 as foci.
 
  • #5
People thanks for the help!

The problem is not supposed to be that hard.

My teacher just said that it was an ellipse. I tried to prove it by the equation.
 

1. How do I identify an ellipse?

An ellipse is a type of conic section that is formed when a plane intersects a cone at an angle that is less than the angle of the cone's base. It can be identified by its curved shape, with two distinct points known as the foci and a major and minor axis.

2. What is the standard equation for an ellipse?

The standard equation for an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) represents the coordinates of the center of the ellipse and a and b represent the lengths of the semi-major and semi-minor axes respectively.

3. How can I determine if an equation is an ellipse?

An equation is an ellipse if it can be put into the standard form mentioned above. This means that it must have both an x-term and a y-term, both with a squared variable, and both with the same coefficients. If the equation does not fit this form, it is not an ellipse.

4. Are there any special cases where an equation is not an ellipse but appears to be?

Yes, there are two special cases where an equation may appear to be an ellipse but is not. The first is when the coefficients of the x-term and y-term are different. This results in an equation for a circle. The second case is when the coefficients of the squared variables are negative, which results in a hyperbola.

5. Can an ellipse have a negative value for either a or b?

Yes, it is possible for an ellipse to have a negative value for either a or b, as long as the other value is positive. This will result in an ellipse that is either horizontally or vertically oriented, depending on which value is negative.

Similar threads

Replies
4
Views
676
Replies
6
Views
1K
Replies
10
Views
1K
  • General Math
Replies
7
Views
722
  • General Math
Replies
3
Views
1K
Replies
2
Views
2K
Replies
1
Views
2K
  • General Math
Replies
8
Views
1K
Replies
15
Views
1K
Replies
4
Views
1K
Back
Top