Question

Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Answer 1

Let P (x, y, z) be any point , the sum of whose distance from the points A(4,0,0) and B($-$4,0,0)
is equal to 10. Then,
PA + PB = 10

$$\begin{gathered} \Rightarrow \sqrt{{\left(x-4\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2}+\sqrt{{\left(x+4\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2}=10 \\ \Rightarrow \sqrt{{\left(x+4\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2}=10-\sqrt{{\left(x-4\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2} \\ \Rightarrow \sqrt{x^2+8x+16+y^2+z^2}=10-\sqrt{x^2-8x+16+y^2+z^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+8x+16+y^2+z^2=100+x^2-8x+16+y^2+z^2-20\sqrt{x^2-8x+16+y^2+z^2} \\ \Rightarrow 16x-100=-20\sqrt{x^2-8x+16+y^2+z^2} \\ \Rightarrow 4x-25=-5\sqrt{x^2-8x+16+y^2+z^2} \\ \Rightarrow 16x^2-200x+625=25\left(x^2-8x+16+y^2+z^2\right) \\ \Rightarrow 16x^2-200x+625=25x^2-200x+400+25y^2+25z^2 \\ \Rightarrow 9x^2+25y^2+25z^2-225=0 \end{gathered}$$

$\therefore 9x^2+25y^2+25z^2-225=0\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{locus}.$