Question

A variable plane at a constant distance p from origin meets the co-ordinate axes in A, B, C. Through these points planes are drawn parallel to co-ordinate planes.Then locus of the point of intersection is

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$
• B. $x^2+y^2+z^2=p^2$
• C. $x+y+z=p$
• D. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=p$

Answer 1

The correct option is A $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Equation of plane is,$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
{a,b,c respectively are intercepts on x,y,z axex}
Then $\frac{abc}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}=p$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}$
Therefore locus of the point (x, y, z) is
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$.