The slopes of the lines given by x^2+2hxy+2y^ | vedclass

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Question

(A) The given equation is $x^2+2hxy+2y^2=0$.Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=h, b=2$.Let the slopes of the lines be $m_1$ and $m_2$.W

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) The given equation is $x^2+2hxy+2y^2=0$.Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=h, b=2$.Let the slopes of the lines be $m_1$ and $m_2$.We know that $m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h$ and $m_1m_2 = \frac{a}{b} = \frac{1}{2}$.Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.Substituting $m_2$ in the product equation: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.If $m_1 = \frac{1}{2}$,then $m_2 = 1$.Then $m_1+m_2 = \frac{1}{2} + 1 = \frac{3}{2}$.Since $m_1+m_2 = -h$,we get $-h = \frac{3}{2} \Rightarrow h = -\frac{3}{2}$.If $m_1 = -\frac{1}{2}$,then $m_2 = -1$.Then $m_1+m_2 = -\frac{1}{2} - 1 = -\frac{3}{2}$.Since $m_1+m_2 = -h$,we get $-h = -\frac{3}{2} \Rightarrow h = \frac{3}{2}$.Thus,$h = \pm \frac{3}{2}$. Given the options,the correct value is $\frac{3}{2}$.

The slopes of the lines given by $x^2+2hxy+2y^2=0$ are in the ratio $1:2$,then $h$ is

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