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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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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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Chapter 28: Introduction to three dimensional coordinate geometry - Exercise 28.2 [Page 10]

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R.D. Sharma Mathematics [English] Class 11
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 17 | Page 10

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