The slopes of the lines represented by x^2+2 | vedclass

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(B) Given the equation of lines: $x^2+2 h x y+2 y^2=0 \dots (i)$ Comparing with $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$. Let the slopes of the

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$\pm \frac{3}{2}$

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(B) Given the equation of lines: $x^2+2 h x y+2 y^2=0 \dots (i)$ Comparing with $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$. Let the slopes of the lines be $m_1$ and $m_2$. Then,$m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h \dots (ii)$ And $m_1 m_2 = \frac{a}{b} = \frac{1}{2} \dots (iii)$ Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$. Substituting $m_2 = 2m_1$ into $(iii)$: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$. If $m_1 = \frac{1}{2}$,then $m_2 = 1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow \frac{1}{2} + 1 = -h$ $\Rightarrow h = -\frac{3}{2}$. If $m_1 = -\frac{1}{2}$,then $m_2 = -1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow -\frac{1}{2} - 1 = -h$ $\Rightarrow h = \frac{3}{2}$. Thus,$h = \pm \frac{3}{2}$.

The slopes of the lines represented by $x^2+2 h x y+2 y^2=0$ are in the ratio $1:2$. Then $h$ is equal to:

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