Question

Show that the point whose distance from $y-axis$ is thrice its distance from $(1,2,-1)$ satisfies the equation $8x^2+9y^2+8z^2-18x-36y+18z+54=0$

Answer 1

Let the points is $P(x,y,z)$

distance from y-axis$=\sqrt{x^2+z^2}$

distance from $(1,2,-1)=\sqrt{(x-1{)}^2+(y-2{)}^2+(z+1{)}^2}$

From the given condition

$\sqrt{x^2+z^2}=3\sqrt{(x-1{)}^2+(y-2{)}^2+(z+1{)}^2}$

squaring and simplifying

$x^2+z^2=9(x^2+y^2+z^2-2x-4y+2z+6)$

$\Rightarrow 0=9x^2+9y^2+9z^2-18x-36y+18z+54-x^2-z^2$

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

$\Rightarrow$ The point satisfy the given condition.