A) \[{{x}^{-1}}+{{y}^{-1}}+{{z}^{-1}}={{p}^{-1}}\]
B) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}={{p}^{-2}}\]
C) \[x+y+z=p\]
D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{p}^{2}}\]
Correct Answer: B
Solution :
[b] Let equation of the variable plane be \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] This meets the coordinate axes at A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Let \[P(\alpha ,\beta ,\gamma )\]be the centroid of the \[\Delta ABC.\] Then \[\alpha =\frac{a+0+0}{3},\beta =\frac{0+b+0}{3},\gamma =\frac{0+0+c}{3}\] \[\therefore a=3\alpha ,b=3\beta ,c=3\gamma \] (2) Plane (1) is at constant distance 3p form the origin, so \[3p=\frac{\left| \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1 \right|}{\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{1}{b} \right)}^{2}}+{{\left( \frac{1}{c} \right)}^{2}}}}\] \[\Rightarrow \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{1}{9{{p}^{2}}}\] (3) Form (2) and (3), we get \[\frac{1}{9{{\alpha }^{2}}}+\frac{1}{9{{\beta }^{2}}}+\frac{1}{9{{\gamma }^{2}}}=\frac{1}{9{{p}^{2}}}\] \[\Rightarrow {{\alpha }^{-2}}+{{\beta }^{-2}}+{{\gamma }^{-2}}={{p}^{-2}}\] Generalizing \[\alpha ,\beta ,\gamma ,\] locus of centroid P \[P(\alpha ,\beta ,\gamma )\]is \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}={{p}^{-2}}\]
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