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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$y^{2} - 2 y - 4 x + 2 z + 6 = 0$

Let the variable point be $P (x, y, z) .$
Its distance from $Y - a x i s = \sqrt{x^{2} + z^{2}}$
Its distance from $A = \sqrt{(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}}$
Given $\sqrt{x + z^{2}} = \sqrt{(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}} \Rightarrow y^{2} - 2 y - 4 x + 2 z + 6 = 0$

Suggest Corrections

Similar questions

The equation of the locus of a point which moves so as to be at equal distances from the point (a, 0) and the $y^{-}$ axis is

The angle between a pair of tangents drawn from a point P to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 s i n^{2} α + 13 c o s^{2} α = 0 i s 2 α .$ The equation of the locus of the point P is

Q. Let P be the point on the parabola,

y^{2} = 8 x

, which is at a minimum distance from the centre C of the circle,

x^{2} + (y + 6)^{2} = 1

. Then, the equation of the circle, passing through C and having its centre at P is

Q. The equation of the circumcircle of the triangle formed by the lines

y + \sqrt{3} x = 6, y - \sqrt{3} x = 6,

and

y = 0,

A point moves in such a way that the sum of square of its distance from the points A(2, 0) and B(-2, 0) is always

equal to the square of the distance between A and B. The locus of the point is

Join BYJU'S Learning Program

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

The correct option is C y2−2y−4x+2z+6=0 Let the variable point be P(x,y,z). Its distance from Y−axis=√x2+z2 Its distance from A=√(x−2)2+(y−1)2+(z+1)2 Given √x+z2=√(x−2)2+(y−1)2+(z+1)2⇒y2−2y−4x+2z+6=0