If operatorname{Tan}^{-1}(2 sqrt{10}) is the | vedclass

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(D) 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 $A = a$,$2H = 4 \implies H = 2$,

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$-\frac{3}{2}$

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(D) 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 $A = a$,$2H = 4 \implies H = 2$,and $B = -2$.Let $	heta$ be the angle between the lines. Then $	an 	heta = \left| \frac{2\sqrt{H^2 - AB}}{A + B} ight|$.Given $	an 	heta = 2\sqrt{10}$,so $2\sqrt{10} = \left| \frac{2\sqrt{2^2 - a(-2)}}{a - 2} ight| = \left| \frac{2\sqrt{4 + 2a}}{a - 2} ight|$.Squaring both sides: $40 = \frac{4(4 + 2a)}{(a - 2)^2} \implies 10 = \frac{4 + 2a}{a^2 - 4a + 4}$.$10a^2 - 40a + 40 = 4 + 2a \implies 10a^2 - 42a + 36 = 0 \implies 5a^2 - 21a + 18 = 0$.Solving for $a$: $5a^2 - 15a - 6a + 18 = 0 \implies 5a(a - 3) - 6(a - 3) = 0 \implies (5a - 6)(a - 3) = 0$.Since $a \in \mathbb{Z}$,we have $a = 3$.The equation of the lines is $3x^2 + 4xy - 2y^2 = 0$. Dividing by $x^2$ and letting $m = y/x$,we get $-2m^2 + 4m + 3 = 0$,or $2m^2 - 4m - 3 = 0$.The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$,which is $-3/2$.

If $\operatorname{Tan}^{-1}(2 \sqrt{10})$ is the angle between the lines $ax^2 + 4xy - 2y^2 = 0$ and $a \in \mathbb{Z}$,then the product of the slopes of the given lines is

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