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

##### 1 Answer

An ellipse.

#### Explanation:

This is the classic way to draw an ellipse.

Attach each end of a string of length

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

Let's do it algebraically:

Suppose

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

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

Subtract

#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

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

Add

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

Divide both sides by

#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

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]}