Question

The equation of the locus of a point whose distance from XY-plane is twice its distance from Z-axis is

• A. \( 2x^2 + 2y^2 - z^2 = 0 \)
• B. \( 2y^2 + 2z^2 - x^2 = 0 \)
• C. \( 4y^2 + 4z^2 - x^2 = 0 \)
• D. \( 4x^2 + 4y^2 - z^2 = 0 \)

Hint:

Memorize the formulas for distances in 3D space: - Distance from P(x,y,z) to XY-plane: \(|z|\) - Distance from P(x,y,z) to YZ-plane: \(|x|\) - Distance from P(x,y,z) to XZ-plane: \(|y|\) - Distance from P(x,y,z) to X-axis: \(\sqrt{y^2+z^2}\) - Distance from P(x,y,z) to Y-axis: \(\sqrt{x^2+z^2}\) - Distance from P(x,y,z) to Z-axis: \(\sqrt{x^2+y^2}\)

Answer 1

The Correct Option is D

Let the point be P(x, y, z).
The distance of a point P(x, y, z) from the XY-plane is the absolute value of its z-coordinate, which is \(|z|\).
The distance of a point P(x, y, z) from the Z-axis is the distance from P to the point (0, 0, z) on the Z-axis. This distance is \( \sqrt{(x-0)^2 + (y-0)^2 + (z-z)^2} = \sqrt{x^2+y^2} \).
The problem states that the distance from the XY-plane is twice the distance from the Z-axis.
So, we have the equation: \( |z| = 2\sqrt{x^2+y^2} \).
To get the equation of the locus, we square both sides to eliminate the absolute value and the square root.
\( (|z|)^2 = (2\sqrt{x^2+y^2})^2 \).
\( z^2 = 4(x^2+y^2) \).
\( z^2 = 4x^2 + 4y^2 \).
Rearranging the terms to match the options, we get:
\( 4x^2 + 4y^2 - z^2 = 0 \).