A) \[{{H}^{2}}\]
B) \[-{{H}^{2}}\]
C) \[2{{H}^{2}}\]
D) \[4{{H}^{2}}\]
Correct Answer: D
Solution :
We know that the pair of lines \[({{a}^{2}}-3{{b}^{2}}){{x}^{2}}+8abxy+({{b}^{2}}-3{{a}^{2}}){{y}^{2}}=0\] with the line \[ax+by+c=0\] form an equilateral triangle. Hence comparing with\[A{{x}^{2}}+2Hxy+B{{y}^{2}}=0\], then \[A={{a}^{2}}-3{{b}^{2}},\,B={{b}^{2}}-3{{a}^{2}},\,2H=8ab\]. Now, \[(A+3B)\,(3A+B)=(-8{{a}^{2}})\,(-8{{b}^{2}})\] = \[{{(8ab)}^{2}}={{(2H)}^{2}}=4{{H}^{2}}\]
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