# Complex Numbers ~ Locus

1. Mar 18, 2009

### Mentallic

1. The problem statement, all variables and given/known data
The curve on the Argand diagram for which
$$|z-2|+|z-4|=10$$
is an ellipse. Find the co-ordinates of its centre and the lengths of its major and minor axes.

2. Relevant equations
$$z=x+iy$$

3. The attempt at a solution
I could only find the centre by using common sense in a way: ($\frac{2+4}{2}$,0) = (3,0)
No idea how to answer the second part though.

2. Mar 18, 2009

### tiny-tim

Hi Mentallic!

Hint: the major axis is obviously along the real axis,

so the ends of the major axis are also, and the ends of the minor axis are on a line parallel to the imaginary axis

3. Mar 18, 2009

### HallsofIvy

Staff Emeritus
You can interpret |z- a| as the distance from z to a. An ellipse has the property that the total distance from a point on the ellipse to the two foci is a constant. In other words |z- 2|+ |z- 4|= 10 tells you that the foci of the ellipse are 2 and 4.

4. Mar 18, 2009

### Mentallic

Yes I realized that too but the distances to these end-points...?

If I were to take a stab at it, I'd go about it like this:

$$\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+y^2}=10$$

where y=0 since end-points are real.

Therefore, $$x-2+x-4=10$$
x=8

So, distance 8 from centres of each modulus.

$$(2-8,0) = (-6,0)$$ and
$$(4+8,0) = (12,0)$$

Is this a legitimate approach?

Similarly for the minor axes:

x=3 since they lie on the perpendicular bisector of (2,0) and (4,0)

Hence, $$\sqrt{1+y^2}+\sqrt{1+y^2}=10$$

$$y=\pm 2\sqrt{6}$$

So I'm guessing the minor end-points are $$(3,2\sqrt{6}) (3,-2\sqrt{6})$$ ?

Once again, this is a total guess so please correct me where I'm wrong.

5. Mar 18, 2009

### Mentallic

So the centre of the ellipse is (3,0) and has focii at (2,0) and (4,0).

So for the general cartesian form of an ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
$$ae=1$$ where e=eccentricity.

I'm lost from here...

6. Mar 18, 2009

### tiny-tim

Hi Mentallic!

(have a square-root: √ )
Yes, and now just say "so 8 is one end of the major axis"

… and then find the other end by using the "opposite value" of the ||s.

(why is that a guess? )

Yes, that's fine, except you mean 3 ±2(√6)i

7. Mar 18, 2009

### HallsofIvy

Staff Emeritus
For an ellipse, $c^2= a^2- b^2$ where a and b are the distances from the center to the vertices and c is the distance from the center to the focus. So you know that $a^2- b^2= 1$.

8. Mar 18, 2009

### Mentallic

Thanks for the correction. I also took on a more geometric approach and found half the minor axis to be $$\sqrt{5^2-1}=2\sqrt{6}$$ through an isosceles triangle and the pythagorean theorem which supported my algebraic result.

Sorry, but I haven't studied ellipses much at all. I remember a few general results but don't know their significance and I'm once again unsure what to do with this relationship of $a^2-b^2=1$

9. Mar 19, 2009

### tiny-tim

Hi Mentallic!

Start by looking at the PF Library pages on conic and eccentricity