If the slope of one of the lines given by ax^ | vedclass

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Question

(A) The given equation is $ax^2+2hxy+by^2=0$. Let the slopes of the lines be $m_1$ and $m_2$. From the properties of homogeneous equations of degree $

Vedclass · Accepted Answer

$8h^2=9ab$

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(A) The given equation is $ax^2+2hxy+by^2=0$. Let the slopes of the lines be $m_1$ and $m_2$. From the properties of homogeneous equations of degree $2$,we have: $m_1+m_2 = -\frac{2h}{b} \dots (1)$ $m_1 \cdot m_2 = \frac{a}{b} \dots (2)$ Given that one slope is twice the other,let $m_1 = 2m_2 \dots (3)$. Substituting $(3)$ into $(1)$: $2m_2 + m_2 = -\frac{2h}{b} \implies 3m_2 = -\frac{2h}{b} \implies m_2 = -\frac{2h}{3b}$. Substituting $(3)$ into $(2)$: $2m_2 \cdot m_2 = \frac{a}{b} \implies 2(m_2)^2 = \frac{a}{b}$. Substituting the value of $m_2$: $2\left(-\frac{2h}{3b}\right)^2 = \frac{a}{b} \implies 2\left(\frac{4h^2}{9b^2}\right) = \frac{a}{b} \implies \frac{8h^2}{9b^2} = \frac{a}{b} \implies 8h^2 = 9ab^2 \cdot \frac{b}{b} = 9ab$. Thus,$8h^2 = 9ab$.

If the slope of one of the lines given by $ax^2+2hxy+by^2=0$ is two times the other,then:

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