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

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

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

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