A) \[{{c}^{2}}=1+{{a}^{2}}\]
B) \[{{c}^{2}}=1-{{a}^{2}}\]
C) \[{{c}^{2}}=1+{{n}^{2}}+{{a}^{2}}\]
D) \[{{c}^{2}}=(1+{{n}^{2}})\,{{a}^{2}}\]
Correct Answer: D
Solution :
The roots of equation will be equal if \[D=0\Rightarrow {{B}^{2}}-4AC=0\] From the equation- \[(1+{{n}^{2}}){{x}^{2}}+2ncx+({{c}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \,\,{{(2nc)}^{2}}-4\times (1+{{n}^{2}})\,({{c}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \,4{{n}^{2}}{{c}^{2}}-(4+4{{n}^{2}})\,({{c}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \,4{{n}^{2}}{{c}^{2}}-4{{c}^{2}}-4{{n}^{2}}{{c}^{2}}+4{{a}^{2}}4{{n}^{2}}{{a}^{2}}=0\] \[\Rightarrow \,4{{a}^{2}}+4{{n}^{2}}{{a}^{2}}-4{{c}^{2}}=0\] \[\Rightarrow \,\,4{{a}^{2}}\,(1+{{n}^{2}})=4{{c}^{2}}\] \[\Rightarrow \,\,\,\]
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