Question

A variable plane is at a distance $p$ from the origin and meet the axes in $A,B,C$. If the locus of the centroid of $\Delta$ $ABC$ $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{k}{p^2}$, then the value of $k$ equals

• A. $4$
• B. $5$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

Let the variable plane be

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

It is at a constant distance $p$ from the origin,

$\therefore p=\frac{1}{\sqrt{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})}}\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}$ ...$\left(1\right)$

The plane cuts axes in $A,B,C$ whose coordinates are $(a,0,0),(0,b,0),(0,0,c).$

Equation of the planes through $A,B,C$ and parallel to coordinates planes are

$x=a$ ...$\left(2\right)$

$y=b$ ...$\left(3\right)$

and, $z=c$ ...$\left(4\right)$

The locus of their point of intesection will be obtained by eliminating $a,b,c$ from these with the help of the relation $\left(1\right)$. We thus get

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$ ,i.e., $x^{-2}+y^{-2}+z^{-2}=p^{-2},$

which is the required locus.

$\therefore k=1$