A) - 16
B) 16
C) 4
D) - 4
Correct Answer: B
Solution :
We have, \[a{{x}^{2}}-{{y}^{2}}+4x-y=0\] Here, \[a=a,\,h=0,\,b=-1;\,f=-\frac{1}{2},\,g=2,\,c=0\] for the straight lines \[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\] \[\Rightarrow \left| \begin{matrix} a & 0 & 2 \\ 0 & -1 & -1/2 \\ 2 & 1/2 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \,a\left[ 0+\left( -\frac{1}{4} \right) \right]-0+2\,[2]=0\] \[\Rightarrow \] \[-\frac{a}{4}+4=0\] or \[a=16\]
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