Question

Show that locus of the point whose
distance from Y-axis is thrice its distance
from (1, 2, -1)
8x2 +9y2 +822-18x-36y +18z+54=0.​

Answer 1

Answer:

To show : 8x^2 +9y^2+ 8z^2 -18x -36y +18z +54=0

here the given point is ( 1,2,-1)

x = 1, y = 2,z = -1

distance from y axis is underroot x^2 + z^2

distance from (1,2,-1) =

sq root of (x-1) ^2 + (y - 2)^2 + (z + 1)^2

given, distance from y axis is thrice the distance from (1,2,-1)

so.,

sq root of x^2 +z^2 = 3 (sq root of (x -1)^2 +(y-2)^2+(z+1)^2)

squaring on both sides we get ,

x^2 + z^2 = 9 [ (x-1)^2 + (y-2)^2 + (z+1)^2 ]

= 9 [x^2 -2x+1 + y^2 -4y +4 +z^2 +2z+1]

= 9x^2 - 18x + 9 + 9y^2 - 36y + 36+ 9 z^2 +18z +9

9x^2 - 18x + 9y^2 - 36y + 9z^2 + 18z + 54 - x^2- z^2 = o

8x^2 +9y^2 +8z^2 -18x -36 y -18 z +54 = 0

hence proved.