A variable plane at a constant distance p fro | vedclass

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(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respective

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.The distance of this plane from the origin $(0, 0, 0)$ is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$.Squaring both sides,we get $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.The points of intersection with the axes are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.Planes drawn through these points parallel to the coordinate planes intersect at the point $P(x, y, z) = (a, b, c)$.Substituting $a=x, b=y, c=z$ into the distance equation,the locus of the point of intersection is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}$.

$A$ variable plane at a constant distance $p$ from the origin meets the coordinate axes at $A, B, C$. Through these points,planes are drawn parallel to the coordinate planes. The locus of the point of intersection is

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