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 pair $ax^

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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 pair $ax^2-7xy-3y^2=0$ shares exactly one line with the second pair,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 $ax^2-7xy-3y^2=0$: $ax^2 - 7x(\frac{2}{3}x) - 3(\frac{2}{3}x)^2 = 0 \implies ax^2 - \frac{14}{3}x^2 - \frac{4}{3}x^2 = 0 \implies ax^2 - 6x^2 = 0$. Thus $a=6$.The equation becomes $6x^2-7xy-3y^2=0$,which factors as $(2x-3y)(3x+y)=0$.Case $2$: If $x+2y=0$ is a factor,then $x = -2y$. Substituting into $ax^2-7xy-3y^2=0$: $a(-2y)^2 - 7(-2y)y - 3y^2 = 0 \implies 4ay^2 + 14y^2 - 3y^2 = 0 \implies 4ay^2 + 11y^2 = 0$. This gives $a = -\frac{11}{4}$,which is not an integer.Thus,$a=6$ and the lines are $2x-3y=0$ and $3x+y=0$.The equation of the bisectors of the angle between $ax^2+2hxy+by^2=0$ is $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.Here $a=6, b=-3, h=-\frac{7}{2}$.$\frac{x^2-y^2}{6-(-3)} = \frac{xy}{-7/2} \implies \frac{x^2-y^2}{9} = -\frac{2xy}{7} \implies 7(x^2-y^2) = -18xy \implies 7x^2+18xy-7y^2=0$.

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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