The gradient of one of the lines x^2 + hxy + | vedclass

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(A) Let the slopes of the lines represented by $x^2 + hxy + 2y^2 = 0$ be $m_1$ and $m_2$. Given that $m_1 = 2m_2$. For the equation $ax^2 + hxy + by^2

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$ \pm 3 $

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(A) Let the slopes of the lines represented by $x^2 + hxy + 2y^2 = 0$ be $m_1$ and $m_2$. Given that $m_1 = 2m_2$. For the equation $ax^2 + hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{h}{b}$ and $m_1 m_2 = \frac{a}{b}$. Here $a = 1$,$h = h$,and $b = 2$. So,$m_1 + m_2 = -\frac{h}{2}$ and $m_1 m_2 = \frac{1}{2}$. Substituting $m_1 = 2m_2$ into the sum: $3m_2 = -\frac{h}{2} \Rightarrow m_2 = -\frac{h}{6}$. Substituting into the product: $(2m_2)(m_2) = \frac{1}{2}$ $\Rightarrow 2m_2^2 = \frac{1}{2}$ $\Rightarrow m_2^2 = \frac{1}{4}$ $\Rightarrow m_2 = \pm \frac{1}{2}$. Now,$3(\pm \frac{1}{2}) = -\frac{h}{2}$ $\Rightarrow \pm \frac{3}{2} = -\frac{h}{2}$ $\Rightarrow h = \mp 3$. Thus,$h = \pm 3$.

The gradient of one of the lines $x^2 + hxy + 2y^2 = 0$ is twice that of the other,then $h =$

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