Question

A variable plane at a constant distance p from the origin meets the axes at A,B,C Through A,B,C planes are drawn parallel to the coordinate planes. Show that the locus of their point of the intersection is given by $x^{-2}+y^{-2}+z^{-2}=p^{-2}$

Answer 1

Let the equation of the variable plane be

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...\left(1\right)$

Now plane (i) is at a constant from $0(0,0,0)$

we get

$P=\frac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$

$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}$

Let the planes pass through A,B,C and parallel to the co-ordinate plane meet is point P then co-ordinate

of P are (a,b,c) To obtain the locum of P, replace a,b,c by x,y,z respectively in (ii) to obtain the required locus as

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$ Or $x^{-2}+y^{-2}+z^{2-}=p^{-2}$