find the equation of the locus of a point whose distance from the y - axis is equal to its distance from (2,1,-1)
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9
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
Answered by
1
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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