The slope of one of the pair of lines 2x^2 + | vedclass

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(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Let the slopes of the two lines be $m$ and $3m$. The equation of the pair of l

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$\pm 8$

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(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Let the slopes of the two lines be $m$ and $3m$. The equation of the pair of lines can be written as $(y - mx)(y - 3mx) = 0$. Expanding this,we get $y^2 - 4mxy + 3m^2x^2 = 0$,or $3m^2x^2 - 4mxy + y^2 = 0$. Dividing the original equation $2x^2 + hxy + 6y^2 = 0$ by $6$,we get $\frac{1}{3}x^2 + \frac{h}{6}xy + y^2 = 0$. Comparing the coefficients of $x^2$ and $xy$ with $3m^2x^2 - 4mxy + y^2 = 0$,we have: $3m^2 = \frac{1}{3}$ $\Rightarrow m^2 = \frac{1}{9}$ $\Rightarrow m = \pm \frac{1}{3}$. Also,$-4m = \frac{h}{6} \Rightarrow h = -24m$. Substituting $m = \pm \frac{1}{3}$ into the expression for $h$: $h = -24(\pm \frac{1}{3}) = \mp 8$. Thus,$h = \pm 8$.

The slope of one of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line. Then,$h = $

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