Answer to Question #250668 in Analytic Geometry for Aliyah

Question #250668

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


1
Expert's answer
2021-10-15T10:00:07-0400

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

Therefore, according to the question:

(x3)2+y2+(x+3)2+y2=8(x3)2+y2=8(x+3)2+y2\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:

(x3)2+y2=64+(x+3)2+y216(x+3)2+y26x=64+6x16(x+3)2+y216(x+3)2+y2=12x+644(x+3)2+y2=3x+16(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+16y2=9x2+256+96x7x2+16y2=112x216+y27=116(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


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