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

Question

The equation of locus of a point whose distance from the y-axis is equal to its distance from the point (2, 1, -1) is (A) x2 + y2 + z2 = 6 (B) x2 - 4x + 2z + 6 = 0 (C) y2 - 2y - 4x + 2z + 6 = 0 (D) x2 + y2 - z2 = 0

Tardigrade Admin · Accepted Answer

The variable point is P (x, y, z). Its distance from the y-axis = √x2 + z2 Its distance from (2, 1, -1) √(x-2)2 +(y-1)2+(z+1)2 Given √x2+z2 = √(x-2)2+(y-1)2+(z+1)2 ⇒ y2 -2y-4x+2z+6=0

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

$x^{2} + y^{2} + z^{2} = 6$

$x^{2} - 4 x + 2 z + 6 = 0$

$y^{2} - 2 y - 4 x + 2 z + 6 = 0$

$x^{2} + y^{2} - z^{2} = 0$

The variable point is P (x, y, z).
Its distance from the y-axis = $x^{2} + z^{2}$
Its distance from (2, 1, -1)
$(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}$
Given
$x^{2} + z^{2} = (x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}$
$\Rightarrow y^{2} - 2 y - 4 x + 2 z + 6 = 0$

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

x2+y2+z2=6

x2−4x+2z+6=0

y2−2y−4x+2z+6=0

x2+y2−z2=0

The variable point is P (x, y, z). Its distance from the y-axis = x2+z2​ Its distance from (2, 1, -1) (x−2)2+(y−1)2+(z+1)2​ Given x2+z2​=(x−2)2+(y−1)2+(z+1)2​ ⇒y2−2y−4x+2z+6=0

$x^{2} + y^{2} + z^{2} = 6$

$x^{2} - 4 x + 2 z + 6 = 0$

$y^{2} - 2 y - 4 x + 2 z + 6 = 0$

$x^{2} + y^{2} - z^{2} = 0$

The variable point is P (x, y, z).
Its distance from the y-axis = $x^{2} + z^{2}$
Its distance from (2, 1, -1)
$(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}$
Given
$x^{2} + z^{2} = (x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}$
$\Rightarrow y^{2} - 2 y - 4 x + 2 z + 6 = 0$