Question

A line segment joining (1, 0, 1) and the origin (0, 0, 0) is revolved about the x-axis to form a right circular cone. If (x, y, z) is any point on the cone other than the origin, then it satisfies the equation

• A. $x^2-2y^2-z^2=0$
• B. $x^2-y^2-z^2=0$
• C. $2x^2-y^2-2z^2=0$
• D. $x^2-2y^2-2z^2=0$

Answer 1

The correct option is B

$x^2-y^2-z^2=0$

Let A be the point (1,0,1) and P(x, y, z) be any point on the cone obtained revolving OA about the x-axis.
From the figure,
$\Delta OPMand\Delta OACaresimilar\frac{OP}{OA}=\frac{OM}{OC}\frac{\sqrt{x^2+y^2+z^2}}{\sqrt{2}}=\frac{x}{1}\Rightarrow x^2+y^2+z^2=2x^2\Rightarrow x^2-y^2-z^2=0$