Q. 175.0( 3 Votes )

# Find the locus of P if PA^{2} + PB^{2} = 2k^{2}, where A and B are the points (3, 4, 5) and (-1, 3, -7).

Answer :

**Given:** Points are A(3, 4, 5) and B(-1, 3, -7)

**To find:** the locus of point P which moves in such a way that PA^{2} + PB^{2} = 2k^{2}

Let the required point P(x, y, z)

**Formula used:**

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(3, 4, 5) is PA,

Distance between P(x, y, z) and B(-1, 3, -7) is PB,

According to question:

PA^{2} + PB^{2} = 2k^{2}

⇒ (x – 3)^{2}+ (y – 4)^{2} + (z – 5)^{2} + (x + 1)^{2} + (y – 3)^{2} + (z + 7)^{2} = 2k^{2}

⇒ x^{2}+ 9 – 6x + y^{2} + 16 – 8y + z^{2} + 25 – 10z + x^{2}+ 1 + 2x + y^{2} + 9 – 6y + z^{2} + 49 + 14z = 2k^{2}

⇒ 2x^{2}+ 2y^{2} + 2z^{2} – 4x – 14y + 4z + 109 = 2k^{2}

⇒ 2x^{2}+ 2y^{2} + 2z^{2} – 4x – 14y + 4z + 109 – 2k^{2} = 0

Hence **locus of point P is 2x ^{2}+ 2y^{2} + 2z^{2} – 4x – 14y + 4z + 109 – 2k^{2} = 0**

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