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(A) Given the equation of the pair of lines is $a^2 x^2 + 2hxy + b^2 y^2 = 0$.Let the slopes of the lines be $m$ and $2m$.From the properties of the h

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$\frac{3 \sqrt{2}}{4}$

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(A) Given the equation of the pair of lines is $a^2 x^2 + 2hxy + b^2 y^2 = 0$.Let the slopes of the lines be $m$ and $2m$.From the properties of the homogeneous equation $Ax^2 + 2Hxy + By^2 = 0$,the sum of slopes is $m_1 + m_2 = -\frac{2H}{B}$ and the product is $m_1 m_2 = \frac{A}{B}$.Here,$A = a^2$,$2H = 2h$,and $B = b^2$.So,$m + 2m = -\frac{2h}{b^2}$ $\Rightarrow 3m = -\frac{2h}{b^2}$ $\Rightarrow m = -\frac{2h}{3b^2}$ $(i)$.Also,$m 	imes 2m = \frac{a^2}{b^2} \Rightarrow 2m^2 = \frac{a^2}{b^2}$ (ii).Substituting $(i)$ into (ii): $2 \left(-\frac{2h}{3b^2}ight)^2 = \frac{a^2}{b^2}$.$2 \left(\frac{4h^2}{9b^4}ight) = \frac{a^2}{b^2} \Rightarrow \frac{8h^2}{9b^4} = \frac{a^2}{b^2}$.$\frac{h^2}{a^2 b^2} = \frac{9}{8} \Rightarrow \frac{h}{ab} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}}$.Rationalizing the denominator: $\frac{3}{2\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4}$.

For $a, b, h > 0$,if the slope of one of the lines represented by $a^2 x^2 + 2hxy + b^2 y^2 = 0$ is twice that of the other,then the value of $\frac{h}{ab}$ is

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