A) -10
B) -9
C) 10
D) 9
Correct Answer: B
Solution :
Given equation is \[2{{x}^{2}}-10xy+12{{y}^{2}}+5x+\lambda y-3=0\] Here,\[a=2,\,\,h=-5,\,\,b=12,\,\,g=\frac{5}{2},\] \[f=\frac{\lambda }{2},\,\,c=-3\] For pair of lines\[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 2 & -5 & 5/2 \\ -5 & 12 & \lambda /2 \\ 5/2 & \lambda /2 & -3 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[2\left( -36-\frac{{{\lambda }^{2}}}{4} \right)+5\left( 15-\frac{5\lambda }{4} \right)\] \[+\frac{5}{2}\left( \frac{-5\lambda }{2}-30 \right)=0\] \[\Rightarrow \]\[-72-\frac{{{\lambda }^{2}}}{2}+75-\frac{25\lambda }{4}-\frac{25\lambda }{4}-75=0\] \[\Rightarrow \] \[{{\lambda }^{2}}+25\lambda +144=0\] \[\Rightarrow \] \[(\lambda +9)(\lambda +16)=0\] \[\Rightarrow \] \[\lambda =-9\] \[(\because \,\,|\lambda |\,\,<16)\]
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