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प्रश्न
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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उत्तर
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
\[\Rightarrow \sqrt{\left( x - 4 \right)^2 + \left( y - 0 \right)^2 + \left( z - 0 \right)^2} + \sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2 + \left( z - 0 \right)^2} = 10\]
\[ \Rightarrow \sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2 + \left( z - 0 \right)^2} = 10 - \sqrt{\left( x - 4 \right)^2 + \left( y - 0 \right)^2 + \left( z - 0 \right)^2}\]
\[ \Rightarrow \sqrt{x^2 + 8x + 16 + y^2 + z^2} = 10 - \sqrt{x^2 - 8x + 16 + y^2 + z^2}\]
\[\Rightarrow x^2 + 8x + 16 + y^2 + z^2 = 100 + x^2 - 8x + 16 + y^2 + z^2 - 20\sqrt{x^2 - 8x + 16 + y^2 + z^2}\]
\[ \Rightarrow 16x - 100 = - 20\sqrt{x^2 - 8x + 16 + y^2 + z^2}\]
\[ \Rightarrow 4x - 25 = - 5\sqrt{x^2 - 8x + 16 + y^2 + z^2}\]
\[ \Rightarrow 16 x^2 - 200x + 625 = 25\left( x^2 - 8x + 16 + y^2 + z^2 \right)\]
\[ \Rightarrow 16 x^2 - 200x + 625 = 25 x^2 - 200x + 400 + 25 y^2 + 25 z^2 \]
\[ \Rightarrow 9 x^2 + 25 y^2 + 25 z^2 - 225 = 0\]
\[\therefore 9 x^2 + 25 y^2 + 25 z^2 - 225 = 0 \text{ is the required locus } .\]
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