A) \[{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}=2\]
B) \[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=2{{c}^{4}}\]
C) \[{{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}=2{{c}^{4}}\]
D) \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0\]
Correct Answer: D
Solution :
Given, \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and \[xy={{c}^{2}}\] \[\therefore \] \[{{x}^{2}}+\left( \frac{{{c}^{2}}}{x} \right)={{a}^{2}}\] \[\Rightarrow \] \[{{x}^{4}}-{{a}^{2}}{{x}^{2}}+{{c}^{4}}=0\] \[\therefore \] \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0\]You need to login to perform this action.
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