Question

find the equation of the locus of a point whose distance from the y - axis is equal to its distance from (2,1,-1)​

Answer 1

Answer:

Any point on Yaxis=0(0,y,,0)=P

Q=(2,1,-1)&A=(x,y,z)

A.T.Q

root of (x-0)2+(y -0)2+(z -0)2=root of (x-2)2+(y -1)2+(z+1)2

x 2 + 0 +z 2 =x 2 +4 -4x+ y 2 +1 -2 y+z 2 +1+2 z

4 x - y2-1+2 y -1 -2 z=4=0

y2-2 y -4 x+ 2 z+6=0

Answer 2

Step-by-step explanation:

Any point on Yaxis=0(0,y,,0)=P

Q=(2,1,-1)&A=(x,y,z)

A.T.Q

root of (x-0)2+(y -0)2+(z -0)2=root of (x-2)2+(y -1)2+(z+1)2

x 2 + 0 +z 2 =x 2 +4 -4x+ y 2 +1 -2 y+z 2 +1+2 z

4 x - y2-1+2 y -1 -2 z=4=0

y2-2 y  -4 x+ 2 z+6=0

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