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]}