If the angle between the pair of lines given | vedclass

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(B) The equation of the pair of lines is $ax^2+4xy+2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=a$,$2H=4 \Rightarrow H=2$,and $B=2$.The a

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$-6 \pm 4\sqrt{3}$

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(B) The equation of the pair of lines is $ax^2+4xy+2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=a$,$2H=4 \Rightarrow H=2$,and $B=2$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^2-AB}}{A+B} ight|$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = 1$.$1 = \left| \frac{2\sqrt{2^2-a(2)}}{a+2} ight| = \left| \frac{2\sqrt{4-2a}}{a+2} ight|$.Squaring both sides,we get $1 = \frac{4(4-2a)}{(a+2)^2}$.$(a+2)^2 = 16-8a \Rightarrow a^2+4a+4 = 16-8a$.$a^2+12a-12 = 0$.Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $a = \frac{-12 \pm \sqrt{144 - 4(1)(-12)}}{2} = \frac{-12 \pm \sqrt{144+48}}{2} = \frac{-12 \pm \sqrt{192}}{2} = \frac{-12 \pm 8\sqrt{3}}{2} = -6 \pm 4\sqrt{3}$.

If the angle between the pair of lines given by the equation $ax^2+4xy+2y^2=0$ is $45^{\circ}$,then the possible values of $a$ are:

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