Question

# Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10.

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Solution

## The given points are A=( 4,0,0 ) and B=( −4,0,0 ) . Let P=( x,y,z ) be a point the sum of whose distances from A and B is 10 units . The formula to find the distance d between two points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is, d= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2  To find the distance AP between the points A and P, the value of x 1 is 4 , x 2 is x , y 1 is 0 , y 2 is y , z 1 is 0 , z 2 is z . Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1), to find the distance AP AP= ( x−4 ) 2 + y 2 + z 2 To find the distance BP between the points B and P, the value of x 1 is −4 , x 2 is x , y 1 is 0 , y 2 is y , z 1 is 0 , z 2 is z . Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1), to find the distance BP AP= ( x+4 ) 2 + y 2 + z 2 It is given that, AP+BP=10 . Substitute values, ( x−4 ) 2 + y 2 + z 2 + ( x+4 ) 2 + y 2 + z 2 =10 ( x−4 ) 2 + y 2 + z 2 =10− ( x+4 ) 2 + y 2 + z 2 Squaring both sides, ( ( x−4 ) 2 + y 2 + z 2 ) 2 = ( 10− ( x+4 ) 2 + y 2 + z 2 ) 2 ( x−4 ) 2 + y 2 + z 2 =100+ ( x+4 ) 2 + y 2 + z 2 −20 ( x+4 ) 2 + y 2 + z 2 x 2 − x 2 −8x−8x+16−16+ y 2 + z 2 − y 2 − z 2 =100−20 ( x+4 ) 2 + y 2 + z 2 100+16x=20 ( x+4 ) 2 + y 2 + z 2 Squaring both sides again, ( 100+16x ) 2 = ( 20 ( x+4 ) 2 + y 2 + z 2 ) 2 ( 25+4x ) 2 = ( 5 ( x+4 ) 2 + y 2 + z 2 ) 2 625+16 x 2 +200x=25 x 2 +200x+400+25 y 2 +25 z 2 9 x 2 +25 y 2 +25 z 2 −225=0 Thus, the equation of the set of points P, the sum of whose distances from the points A=( 4,0,0 ) and B=( −4,0,0 ) is 10 units is 9 x 2 +25 y 2 +25 z 2 −225=0 .

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