What is the Locus of Complex Numbers in an Ellipse?

Mentallic
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Homework Statement


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.


Homework Equations


z=x+iy


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.
 
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Mentallic said:
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.

Hi Mentallic! :smile:

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 :wink:
 
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.
 
tiny-tim said:
Hi Mentallic! :smile:

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 :wink:

Yes I realized that too :smile: 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.
 
HallsofIvy said:
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.

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...
 
Hi Mentallic! :smile:

(have a square-root: √ :wink:)
Mentallic said:
\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

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.

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

(why is that a guess? :confused:)

Yes, that's fine, except you mean 3 ±2(√6)i :smile:
 
Mentallic said:
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...
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.
 
tiny-tim said:
Hi Mentallic! :smile:

(have a square-root: √ :wink:)


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? :confused:)

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

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.

HallsofIvy said:
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.
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
 
Mentallic said:
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

Hi Mentallic! :smile:

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