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Question
Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).
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Solution
Let P (x, y, z) be the point if \[P A^2 + P B^2 = 2 k^2\]
\[\Rightarrow \left( \sqrt{\left( x - 3 \right)^2 + \left( y - 4 \right)^2 + \left( z - 5 \right)^2} \right)^2 + \left( \sqrt{\left( x + 1 \right)^2 + \left( y - 3 \right)^2 + \left( z + 7 \right)^2} \right)^2 = 2 k^2 \]
\[ \Rightarrow x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 - 10z + 25 + x^2 + 2x + 1 + y^2 - 6y + 9 + z^2 + 14z + 49 = 2 k^2 \]
\[ \Rightarrow 2 x^2 + 2 y^2 + 2 z^2 - 4x - 14y + 4z + 109 - 2 k^2 = 0\]
Hence, \[2 x^2 + 2 y^2 + 2 z^2 - 4x - 14y + 4z + 109 - 2 k^2 = 0\] is the locus of point P.
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