If the equation ax^2+2hxy+by^2+2gx+2fy+c=0 re | vedclass

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(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. If this represents two straight lines,the distance of these lines f

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$c(bf^2-ag^2)$

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(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. If this represents two straight lines,the distance of these lines from the origin $(0,0)$ is given by $d = \frac{|c_i|}{\sqrt{1+m_i^2}}$.Given that the lines are equidistant from the origin,we have $\frac{|c_1|}{\sqrt{1+m_1^2}} = \frac{|c_2|}{\sqrt{1+m_2^2}}$,which implies $c_1^2(1+m_2^2) = c_2^2(1+m_1^2)$.Rearranging gives $c_1^2-c_2^2 = c_2^2m_1^2-c_1^2m_2^2$,or $(c_1+c_2)(c_1-c_2) = (c_2m_1+c_1m_2)(c_2m_1-c_1m_2)$.Using the properties of the pair of straight lines,we know $c_1+c_2 = -\frac{2f}{b}$,$c_1c_2 = \frac{c}{b}$,$m_1+m_2 = -\frac{2h}{b}$,$m_1m_2 = \frac{a}{b}$,and $m_1c_2+m_2c_1 = \frac{2g}{b}$.Substituting these values into the condition,we get $-\frac{2f}{b} \cdot \sqrt{(\frac{2f}{b})^2 - 4\frac{c}{b}} = \frac{2g}{b} \cdot \sqrt{(\frac{2g}{b})^2 - 4\frac{a}{b}}$.Squaring both sides leads to $f^2(f^2-bc) = g^2(g^2-ac)$.This simplifies to $f^4-f^2bc = g^4-g^2ac$,which rearranges to $f^4-g^4 = c(bf^2-ag^2)$.

If the equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents two straight lines equidistant from the origin,then $f^4-g^4=$

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