A) \[{{x}^{2}}+{{a}^{2}}=2{{c}^{2}}-{{y}^{2}}-{{z}^{2}}\]
B) \[{{x}^{2}}+{{a}^{2}}={{c}^{2}}-{{y}^{2}}-{{z}^{2}}\]
C) \[{{x}^{2}}-{{a}^{2}}={{c}^{2}}-{{y}^{2}}-{{z}^{2}}\]
D) \[{{x}^{2}}+{{a}^{2}}={{c}^{2}}+{{y}^{2}}+{{z}^{2}}\]
Correct Answer: B
Solution :
[b] Let the point be P(x, y, z) and two points, (a, 0, 0) and (-a, 0, 0) be A and B As given in the problem, \[P{{A}^{2}}+P{{B}^{2}}=2{{c}^{2}}\] So, \[{{(x+a)}^{2}}+{{(y-0)}^{2}}+{{(z-0)}^{2}}\] \[+{{(x-a)}^{2}}+{{(y-0)}^{2}}+{{(z-0)}^{2}}=2{{c}^{2}}\] or, \[{{(x+a)}^{2}}+{{y}^{2}}+{{z}^{2}}+{{(x-a)}^{2}}+{{y}^{2}}+{{z}^{2}}=2{{c}^{2}}\] \[{{x}^{2}}+2a+{{a}^{2}}+{{y}^{2}}+{{z}^{2}}+{{x}^{2}}-2a+{{a}^{2}}+{{y}^{2}}+{{z}^{2}}=2{{c}^{2}}\]\[\Rightarrow 2({{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{a}^{2}})=2{{c}^{2}}\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{a}^{2}}={{c}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}={{c}^{2}}-{{y}^{2}}-{{z}^{2}}\]You need to login to perform this action.
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