Question

What is the locus of points `z` in the complex plane satisfying `abs(z-3i) + abs(z+3i) = 10` ?

Answer 1

An ellipse.

Explanation:

This is the classic way to draw an ellipse.

Attach each end of a string of length `10` to two pegs - one at `3i` and the other at `-3i` (`6` units apart). These are the focii of the ellipse.

Take a pencil and pull the string tight, then move the pencil around the pegs, allowing the string to slide past the end of the pencil, but keeping it taught.

The resulting shape will be an ellipse with major axis running from `5i` to `-5i` and minor axis from `-4` to `4`.

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Let's do it algebraically:

Suppose `z = x+yi`

Then:

`10 = abs(x+(y-3)i) + abs(x+(y+3i))`

`= sqrt(x^2+(y-3)^2)+sqrt(x^2+(y+3)^2)`

`= sqrt(x^2+y^2-6y+9)+sqrt(x^2+y^2+6y+9)`

Squaring both ends, we get:

`100 = (x^2+y^2-6y+9) + (x^2+y^2+6y+9) + 2sqrt(x^2+y^2-6y+9)sqrt(x^2+y^2+6y+9)`

`= (2x^2+2y^2+18) + 2sqrt((x^2+y^2-6y+9)(x^2+y^2+6y+9))`

Divide both ends by `2` to get:

`50 = (x^2+y^2+9)+sqrt((x^2+y^2-6y+9)(x^2+y^2+6y+9))`

Subtract `(x^2+y^2+9)` from both sides to get:

`41-x^2-y^2 = sqrt((x^2+y^2-6y+9)(x^2+y^2+6y+9))`

Square both sides to get:

`x^4+2x^2y^2-82x^2+y^4-82y^2+1681`

`=(x^2+y^2-6y+9)(x^2+y^2+6y+9)`

`=x^4+2x^2y^2+18x^2+y^4-18y^2+81`

Subtract `x^4+2x^2y^2+y^4` from both ends to get:

`-82x^2-82y^2+1681=18x^2-18y^2+81`

Add `82x^2+82y^2-81` to both sides to get:

`1600=100x^2+64y^2`

Divide both sides by `1600` and transpose to get:

`x^2/4^2+y^2/5^2 = 1`

which is the standard form of the equation of an ellipse with horizontal semi minor axis `4` and vertical semi major axis `5`.

graph{(x^2/16+y^2/25-1)(x^2+(y-3)^2-0.02)(x^2+(y+3)^2-0.02)=0 [-10, 10, -5.6, 5.6]}