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(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Dividing by $x^2$,we get $b(y/x)^2 + 2h(y/x) + a = 0$. Let $m_1$ and $m_2$ be the slopes of the li

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$ax^2 - 2hxy + by^2 = 0$

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(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Dividing by $x^2$,we get $b(y/x)^2 + 2h(y/x) + a = 0$. Let $m_1$ and $m_2$ be the slopes of the lines,so $m_1 + m_2 = -2h/b$ and $m_1m_2 = a/b$. The reflection of a line $y = mx$ in the line $y = 0$ (the $x$-axis) is $y = -mx$. Thus,the new slopes are $-m_1$ and $-m_2$. The new equation is $(y - (-m_1)x)(y - (-m_2)x) = 0$,which is $(y + m_1x)(y + m_2x) = 0$. Expanding this,we get $y^2 + (m_1 + m_2)xy + m_1m_2x^2 = 0$. Substituting the values,$y^2 - (2h/b)xy + (a/b)x^2 = 0$. Multiplying by $b$,we get $by^2 - 2hxy + ax^2 = 0$,which is $ax^2 - 2hxy + by^2 = 0$.

The image of the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ by the line mirror $y = 0$ is

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