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.

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Solution

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 $⇒\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}+\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}=10\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}=10-\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}+{\left(z-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}+8x+16+{y}^{2}+{z}^{2}}=10-\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $⇒{x}^{2}+8x+16+{y}^{2}+{z}^{2}=100+{x}^{2}-8x+16+{y}^{2}+{z}^{2}-20\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒16x-100=-20\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒4x-25=-5\sqrt{{x}^{2}-8x+16+{y}^{2}+{z}^{2}}\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-200x+625=25\left({x}^{2}-8x+16+{y}^{2}+{z}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-200x+625=25{x}^{2}-200x+400+25{y}^{2}+25{z}^{2}\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}+25{y}^{2}+25{z}^{2}-225=0$ $\therefore 9{x}^{2}+25{y}^{2}+25{z}^{2}-225=0\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{locus}.$

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