Question

The locus of the point,which moves in such a way that the distance from the origin is thrice the distance from the xy plane .find the equation......(need of the hour)....

Answer 1

Let the point be (x,y,z)

Distance from xy plane = z

Distance from origin = $\sqrt[2]{ x^{2} + z^{2} + y^{2} }$

Given , $\sqrt[2]{ x^{2} + z^{2} + y^{2} }$ = 3z

So , x^2 + y^2 + z^2 = 9z^2

x^2 + y^2 = 8z^2

Locus is x^2 + y ^2 - 8 z^2 = 0

Answer 2

Answer:

The equation of the locus of the point P is x² + y² = 8z².

Step-by-step explanation:

Let the point be P(x, y, z) and O(0, 0, 0) be the origin.

According to the given information, the distance of point P from the origin is thrice its distance from the xy plane. This can be written as:

√(x² + y² + z²) = 3|z|

Squaring both sides of the equation, we get:

x² + y² + z² = 9z²

x² + y² = 8z²

This is the equation of a cone whose vertex is at the origin and axis is along the z-axis. The base of the cone is a circle with radius √8 times the height.

Thus, the equation of the locus of the point P is x² + y² = 8z².

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