Representation of an ellipse equation

Question

Question: Show that $\vert z-4i\vert$ + $\vert z+4i\vert$=$10$ represents the equation of an ellipse
Definition of an ellipse: $\frac{x^2}{a^2}$+$\frac{y^2}{b^2}$=$1$ where $a$ and $b$ are positive numbers, is called an ellipse in standard position.
Foci:Two fixed pionts.
Hint:Use the definition of an ellipse as the locus of points whose sum of distances from two fixed foci is a constant

The question itself is confusing me. I need help

Answer 1

Here's a soul crushing method:

We have $$|z-4i| +|z+4i|=10$$

Let $$z= x +iy$$

Then:

$$|x+iy-4i|+|x+iy+4i|=10$$

$$|x+(y-4)i|+|x+(y+4)i|=10$$

$$\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10$$

$$x^2+(y-4)^2+x^2+(y+4)^2+2\sqrt{x^2+(y-4)^2}\sqrt{x^2+(y+4)^2} = 100$$

$$2x^2+2y^2+32 -100 = -2\sqrt{x^2+(y-4)^2}\sqrt{x^2+(y+4)^2}$$

$$x^2+y^2- 34 = -\sqrt{x^2+(y-4)^2}\sqrt{x^2+(y+4)^2}$$

$$(x^2+y^2-34)^2 = (x^2+(y-4)^2)(x^2+(y+4)^2)$$

$$x^4+x^{2}y^{2}-34x^2+x^2y^2+y^4-34y^2-34x^2-34y^2+1156 = x^4+x^2[(y+4)^2+(y-4)^2]+(y-4)^2(y+4)^2$$

$$x^4+2x^2y^2-64x^2-64y^2+y^4+1156=x^4+x^2(2y^2+32)+y^4-32y^2+256$$

$$\require{cancel}\cancel{x^4}+\cancel{2x^2y^2}-64x^2-64y^2+\cancel{y^4}+1156=\cancel{x^4}+\cancel{2x^{2}y^2}+32x^2+\cancel{y^4}-32y^2+256$$

$$32y^2+96x^2= 900$$

$$\frac{8y^2}{225}+\frac{8x^2}{75} = 1$$

Let $a^2 = \frac{75}{8} \Rightarrow a= \frac{5\sqrt{6}}{4}$

Let $b^2 = \frac{225}{8}\Rightarrow b =\frac{15\sqrt{2}}{4}$

Then:

$$\frac{x^2}{(\frac{5\sqrt{6}}{4})^2} + \frac{y^2}{(\frac{15\sqrt{2}}{4})^2} = 1$$ which is in the form of an ellipse as required.