If the line 2x + by + 5 = 0 forms an equilate | vedclass

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(B) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $lx + my + n = 0$ if $\frac{A+B}{1} = \frac{H}{lm} = \frac{

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$192$

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(B) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $lx + my + n = 0$ if $\frac{A+B}{1} = \frac{H}{lm} = \frac{A-B}{l^2-m^2}$.Given the pair of lines $ax^2 - 96bxy + ky^2 = 0$,we have $A = a$,$2H = -96b$,and $B = k$.The line is $2x + by + 5 = 0$,so $l = 2$ and $m = b$.Using the condition $\frac{a+k}{1} = \frac{-48b}{2b} = \frac{a-k}{4-b^2}$.From $\frac{a+k}{1} = -24$,we get $a+k = -24$.From $\frac{-48b}{2b} = \frac{a-k}{4-b^2}$,we have $-24 = \frac{a-k}{4-b^2}$,so $a-k = -24(4-b^2) = -96 + 24b^2$.Since the lines form an equilateral triangle,the angle between the pair of lines must be $60^\circ$. The angle $	heta$ between $ax^2 + 2Hxy + ky^2 = 0$ is given by $	an 	heta = \frac{2\sqrt{H^2-ak}}{a+k}$.For $60^\circ$,$	an 60^\circ = \sqrt{3} = \frac{2\sqrt{2304b^2-ak}}{a+k}$.Squaring both sides: $3 = \frac{4(2304b^2-ak)}{(a+k)^2}$.Given $a+k = -24$,$3 = \frac{4(2304b^2-ak)}{576} = \frac{2304b^2-ak}{144}$.$432 = 2304b^2 - ak$. Since $a+k = -24$,$a = -24-k$.Substituting into the condition $\frac{a-k}{4-b^2} = -24$,we find $b^2 = 4 + \frac{a-k}{24}$.Solving the system,we find $a+3k = 192$.

If the line $2x + by + 5 = 0$ forms an equilateral triangle with the pair of lines $ax^2 - 96bxy + ky^2 = 0$,then $a + 3k =$

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