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(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Let the lines be $y = m_1x$ and $y = m_2x$. Then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a

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

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(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Let the lines be $y = m_1x$ and $y = m_2x$. Then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$. The image of a line $y = mx$ in the mirror $y = 0$ (the $x$-axis) is $y = -mx$. Thus,the new lines are $y = -m_1x$ and $y = -m_2x$. The combined equation 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,we get $y^2 + (-\frac{2h}{b})xy + (\frac{a}{b})x^2 = 0$. 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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