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in Mathematics by (52.7k points)

Find the equation of set of points P such that PA2  + PB2  = 2k2 , where A and B are the points (3,4,5) and (-1,3, -7), respectively.

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Let P = (x,y,z),B = (3,4,5) and B = (-1,3-7). 

Now, PB2  = (x +1)2  + (y – 3)2  + (z + 7)2  and 

PA2 = (x – 3)2  + (y – 4)2  + (z – 5)2 

Given that PA2  + PB2  = 2k2, then 

(x - 3)2  + (y - 4)2  +(z - 5)2  + (x + 1)2 +(y - 3)+ (z + 7)2  = 2k2

⇒ x2  – 62+ 9 + y2  – 8y + 16 + z2  – 10z + 22 + x2  + 2x +1 + y2  – 6y + 9 + z2  +14z + 49 = 2k. 

⇒ 2x2  + 2y2  + 2z2  – 4x – 14y + 4z + 109 = 2k2 

⇒ 2x2  + 2y2  + 2z2  – Ax – 14y + 4z = 2k2  -109

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