If the pair of lines ax^2-7xy-3y^2=0 and 2x^2 | vedclass

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(A) The second pair of lines is $2x^2+xy-6y^2=0$. Factoring this,we get $(2x-3y)(x+2y)=0$. So the lines are $2x-3y=0$ and $x+2y=0$. Since the two pair

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$7x^2+18xy-7y^2=0$

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(A) The second pair of lines is $2x^2+xy-6y^2=0$. Factoring this,we get $(2x-3y)(x+2y)=0$. So the lines are $2x-3y=0$ and $x+2y=0$. Since the two pairs of lines have exactly one line in common,either $2x-3y=0$ or $x+2y=0$ must be a factor of $ax^2-7xy-3y^2=0$. Case $1$: If $2x-3y=0$ is a factor,then $y = \frac{2}{3}x$. Substituting into $ax^2-7xy-3y^2=0$,we get $ax^2-7x(\frac{2}{3}x)-3(\frac{4}{9}x^2)=0$,which simplifies to $ax^2 - \frac{14}{3}x^2 - \frac{4}{3}x^2 = 0$,so $a - 6 = 0$,giving $a=6$. Case $2$: If $x+2y=0$ is a factor,then $x = -2y$. Substituting into $ax^2-7xy-3y^2=0$,we get $a(4y^2)-7(-2y)y-3y^2=0$,which simplifies to $4ay^2 + 14y^2 - 3y^2 = 0$,so $4a + 11 = 0$,giving $a = -\frac{11}{4}$,which is not an integer. Thus,$a=6$. The equation is $6x^2-7xy-3y^2=0$. The equation of the bisectors of the angles between $Ax^2+2Hxy+By^2=0$ is $\frac{x^2-y^2}{A-B} = \frac{xy}{H}$. Here $A=6, H=-\frac{7}{2}, B=-3$. So,$\frac{x^2-y^2}{6-(-3)} = \frac{xy}{-7/2} \implies \frac{x^2-y^2}{9} = -\frac{2xy}{7}$. $7(x^2-y^2) = -18xy \implies 7x^2+18xy-7y^2=0$.

If the pair of lines $ax^2-7xy-3y^2=0$ and $2x^2+xy-6y^2=0$ have exactly one line in common and $a$ is an integer,then the equation of the pair of bisectors of the angles between the lines $ax^2-7xy-3y^2=0$ is

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