A variable plane at a constant distance p fro | vedclass

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Question

(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 $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$.Since the planes are drawn parallel to the coordinate planes through $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the point of intersection of these planes is $(a, b, c)$.Replacing $(a, b, c)$ with $(x, y, z)$,the locus 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 points $A, B, C$. Through these points,planes are drawn parallel to the coordinate planes. Find the locus of their point of intersection.

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