Ellipse Complex Numbers Equation $|z - 3| + |z +3| = 10$

Question

How do I show algebraically that $$|z - 3| + |z +3| = 10$$

where $z = x + iy$ (x and y are real) is equivalent to $$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$$

I know it's easy to see that the equation is an ellipse but I'm still curious to know to prove this algebraically. I tried squaring both sides but the equation didn't simplify.

Answer 1

Let $z=x+iy$. Then, $|z + 3|^2 = (10 -|z -3|)^2$ leads to

$$(x + 3+iy)(x + 3-iy) = 100 + (x - 3+iy)(x - 3-iy)-20 |z -3|$$

which reduces to

$$25-3x = 5|z -3|$$

or

$$(25-3x)^2 = 25(x -3+iy)(x -3-iy)=25(x^2-6x+9+y^2)$$

Simplify to get

$$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$$

Answer 2

It is $$\sqrt{(x-3)^2+y^2}+\sqrt{(x+3)^2+y^2}=10$$ After squaring and simplifying we get $$3x+25=5\sqrt{(x+3)^2+y^2}$$ Squaring again and combining like terms we get $$400=16x^2+25y^2$$