If the equation ax^2 + 2hxy + by^2 = 0 has on | vedclass

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(D) The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines passing through the origin. Let the slopes of the two lines be $m_1$ and $m_2$. W

Vedclass · Accepted Answer

$(a + b)^2 = 4h^2$

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(D) The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines passing through the origin. Let the slopes of the two lines be $m_1$ and $m_2$. We know that $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$. The bisectors of the angles between the coordinate axes are the lines $y = x$ and $y = -x$,which have slopes $1$ and $-1$ respectively. Case $1$: Let $m_1 = 1$. Then $m_2 = \frac{a}{b}$. From $m_1 + m_2 = -\frac{2h}{b}$,we get $1 + \frac{a}{b} = -\frac{2h}{b}$ $\Rightarrow \frac{a+b}{b} = -\frac{2h}{b}$ $\Rightarrow a+b = -2h$. Squaring both sides,we get $(a+b)^2 = 4h^2$. Case $2$: Let $m_1 = -1$. Then $m_2 = -\frac{a}{b}$. From $m_1 + m_2 = -\frac{2h}{b}$,we get $-1 - \frac{a}{b} = -\frac{2h}{b}$ $\Rightarrow \frac{a+b}{b} = \frac{2h}{b}$ $\Rightarrow a+b = 2h$. Squaring both sides,we get $(a+b)^2 = 4h^2$. Thus,in both cases,the condition is $(a+b)^2 = 4h^2$.

If the equation $ax^2 + 2hxy + by^2 = 0$ has one line as the bisector of the angle between the coordinate axes,then:

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