If m_1 and m_2 are the slopes of the lines re | vedclass

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(B) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. Dividing by $x^2$,we get $b(y/x)^2 + 2h(y/x) + a = 0$. Let $m = y/x$,so $bm^2 + 2hm

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$m_1 = 4m_2$

Vedclass · Answer

(B) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. Dividing by $x^2$,we get $b(y/x)^2 + 2h(y/x) + a = 0$. Let $m = y/x$,so $bm^2 + 2hm + a = 0$. The roots are $m_1$ and $m_2$,so $m_1 + m_2 = -2h/b$ and $m_1m_2 = a/b$. The difference of slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2} = \sqrt{(-2h/b)^2 - 4(a/b)} = \sqrt{(4h^2 - 4ab)/b^2} = \frac{2}{b} \sqrt{h^2 - ab}$. Given $16h^2 = 25ab$,we have $h^2 = \frac{25}{16}ab$. Substituting this into the difference formula: $|m_1 - m_2| = \frac{2}{|b|} \sqrt{\frac{25}{16}ab - ab} = \frac{2}{|b|} \sqrt{\frac{9}{16}ab} = \frac{2}{|b|} \cdot \frac{3}{4} \sqrt{ab} = \frac{3}{2} \sqrt{\frac{a}{b}}$. However,checking the options provided,if we assume the question implies a specific ratio or relationship,let's re-evaluate. If $16h^2 = 25ab$,then $h^2/ab = 25/16$. $|m_1 - m_2| = \frac{2}{|b|} \sqrt{h^2 - ab} = \frac{2}{|b|} \sqrt{\frac{25}{16}ab - ab} = \frac{2}{|b|} \sqrt{\frac{9}{16}ab} = \frac{3}{2} \sqrt{\frac{a}{b}}$. Given the standard nature of such problems,if $m_1 = 4m_2$ or similar,we test $m_1 = 4m_2$. Then $m_1+m_2 = 5m_2 = -2h/b$ and $m_1m_2 = 4m_2^2 = a/b$. $m_2^2 = a/4b$,so $m_2 = \frac{1}{2}\sqrt{a/b}$. $5m_2 = \frac{5}{2}\sqrt{a/b} = -2h/b \implies \frac{25}{4}(a/b) = 4h^2/b^2 \implies 25a = 16h^2/b \implies 25ab = 16h^2$. This matches the given condition. Thus,$m_1 = 4m_2$ is correct.

If $m_1$ and $m_2$ are the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ satisfying the condition $16h^2 = 25ab$,then $\ldots$.

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