A) \[{{a}^{2}}-{{b}^{2}}+2ac=0\]
B) \[{{(a-c)}^{2}}={{b}^{2}}+{{c}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}-2ac=0\]
D) \[{{a}^{2}}+{{b}^{2}}+2ac=0\]
Correct Answer: A
Solution :
As given, \[\sin \alpha +\cos \alpha =-\frac{b}{a},\]\[\sin \alpha \cos \alpha =\frac{c}{a}\] To eliminate\[\alpha \], we have \[1={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha ={{(\sin \alpha +\cos \alpha )}^{2}}-2\sin \alpha \cos \alpha \] \[=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}\,\,\Rightarrow {{a}^{2}}-{{b}^{2}}+2ac=0\]
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