Question

A variable plane is at a constant distance $3p$ from the origin and meets the axes in $A,B$ and $C$. The locus of the centroid of the triangle $ABC$ is

• A. $x^{-2}+y^{-2}+z^{-2}=p^{-2}$
• B. $x^{-2}+y^{-2}+z^{-2}=3p^{-2}$
• C. $x^{-2}+y^{-2}+z^{-2}=9p^{-2}$
• D. $x^{-2}+y^{-2}+z^{-2}=16p^{-2}$

Answer 1

The correct option is A $x^{-2}+y^{-2}+z^{-2}=p^{-2}$

Equation of plane at a distance $3p$ from origin is given by,
$lx+my+nz=3p$, where $l,m,n$ are direction cosine of normal to the plane along distance.
Thus intercept on the axes are,
$A=(\frac{3p}{l},0,0),B=(0,\frac{3p}{m},0),C=(0,0,\frac{3p}{n})$

So centroid of triangle $ABC$ is,
$x=\frac{p}{l},y=\frac{p}{m},z=\frac{p}{n}$

Also we know,
$l^2+m^2+n^2=1$
$\Rightarrow \frac{p^2}{x^2}+\frac{p^2}{y^2}+\frac{p^2}{z^2}=1$

Hence, required locus is
$x^{-2}+y^{-2}+z^{-2}=p^{-2}$