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 $2x^2+4xy-3xy-6y^2=0$,which is $(2x-3y)(x+2y)=0$. The lines are $2x-3y=0$ and

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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 $2x^2+4xy-3xy-6y^2=0$,which is $(2x-3y)(x+2y)=0$. The lines are $2x-3y=0$ and $x+2y=0$. Since the first pair $ax^2-7xy-3y^2=0$ shares exactly one line with the second,either $2x-3y=0$ or $x+2y=0$ must be a factor of $ax^2-7xy-3y^2=0$. If $2x-3y=0$ is a factor,then $y = \frac{2}{3}x$. Substituting into $ax^2-7x(\frac{2}{3}x)-3(\frac{2}{3}x)^2=0$,we get $ax^2 - \frac{14}{3}x^2 - \frac{4}{3}x^2 = 0$,so $a - 6 = 0$,which gives $a=6$. If $x+2y=0$ is a factor,then $x = -2y$. Substituting into $a(-2y)^2-7(-2y)y-3y^2=0$,we get $4ay^2 + 14y^2 - 3y^2 = 0$,so $4a + 11 = 0$,which gives $a = -\frac{11}{4}$ (not an integer). Thus,$a=6$. The equation is $6x^2-7xy-3y^2=0$. The equation of the bisectors of the angle between $Ax^2+Bxy+Cy^2=0$ is $\frac{x^2-y^2}{A-C} = \frac{xy}{B/2}$. Here $A=6, B=-7, C=-3$. So $\frac{x^2-y^2}{6-(-3)} = \frac{xy}{-7/2} \Rightarrow \frac{x^2-y^2}{9} = \frac{-2xy}{7}$. $7(x^2-y^2) = -18xy \Rightarrow 7x^2+18xy-7y^2=0$.

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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