A) \[7/2\]
B) \[-19\]
C) \[-12\]
D) \[12\]
Correct Answer: C
Solution :
\[12{{x}^{2}}+7xy+a{{y}^{2}}+13x-y+3=0\] \[a=2,\text{ }b=a,\]for perpendiculaty coeff. of \[{{x}^{2}}+\]coeff. of \[{{y}^{2}}=0\] \[a+b=0\Rightarrow 12+a=0\Rightarrow a=-12\]
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