If the pair of lines 6x^2+xy-y^2=0 and 3x^2-a | vedclass

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(A) The given pair of lines is $6x^2+xy-y^2=0$. Factorizing the equation: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightarrow (3x-y)(2x

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$\frac{1}{2}$

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(A) The given pair of lines is $6x^2+xy-y^2=0$. Factorizing the equation: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightarrow (3x-y)(2x+y)=0$. Thus,the lines are $3x-y=0$ and $2x+y=0$. Case $1$: If $3x-y=0$ is a common line,then $y=3x$. Substituting $y=3x$ into $3x^2-axy-y^2=0$: $3x^2-ax(3x)-(3x)^2=0$ $\Rightarrow 3x^2-3ax^2-9x^2=0$ $\Rightarrow -3ax^2=6x^2$ $\Rightarrow a=-2$. Since $a>0$,this case is rejected. Case $2$: If $2x+y=0$ is a common line,then $y=-2x$. Substituting $y=-2x$ into $3x^2-axy-y^2=0$: $3x^2-ax(-2x)-(-2x)^2=0$ $\Rightarrow 3x^2+2ax^2-4x^2=0$ $\Rightarrow 2ax^2=x^2$ $\Rightarrow 2a=1$ $\Rightarrow a=\frac{1}{2}$. Thus,$a=\frac{1}{2}$.

If the pair of lines $6x^2+xy-y^2=0$ and $3x^2-axy-y^2=0$ with $a>0$ have a common line,then $a=$

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