Question

The sum of the distances of a moving point from (3, 0) and (-3, 0) is 8. Find its equation

Answer 1

Let the coordinates of the moving point be $(x,y).$

Therefore, according to the question:

$\sqrt{(x-3)^2+y^2}+\sqrt{(x+3)^2+y^2}=8\\ \Rightarrow \sqrt{(x-3)^2+y^2}=8-\sqrt{(x+3)^2+y^2}$

Squaring both sides, we get:

$(x-3)^2+y^2=64+(x+3)^2+y^2-16\sqrt{(x+3)^2+y^2}\\ \Rightarrow -6x=64+6x-16\sqrt{(x+3)^2+y^2}\\ \Rightarrow 16\sqrt{(x+3)^2+y^2}=12x+64\\ \Rightarrow 4\sqrt{(x+3)^2+y^2}=3x+16$

Squaring both sides, we get:

$16(x+3)^2+16y^2=9x^2+256+96x\\ \Rightarrow 7x^2+16y^2=112\\ \Rightarrow \frac{x^2}{16}+\frac{y^2}{7}=1$