If one of the lines represented by ax^2+2hxy+ | vedclass

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(B) The given pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $a+2h(\frac{y}{x})+b(\frac{y}{x})^2=0$. Let $k = \frac{y}{x}$ be the slope

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$am^2+2hmn+bn^2=0$

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(B) The given pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $a+2h(\frac{y}{x})+b(\frac{y}{x})^2=0$. Let $k = \frac{y}{x}$ be the slope of one of the lines. Then $bk^2+2hk+a=0$. The slope of the line $mx+ny=18$ is $-\frac{m}{n}$. Since the line is perpendicular to $mx+ny=18$,its slope $k$ must be the negative reciprocal of $-\frac{m}{n}$,so $k = \frac{n}{m}$. Substituting $k = \frac{n}{m}$ into the equation $bk^2+2hk+a=0$,we get $b(\frac{n}{m})^2+2h(\frac{n}{m})+a=0$. Multiplying by $m^2$,we obtain $bn^2+2hmn+am^2=0$.

If one of the lines represented by $ax^2+2hxy+by^2=0$ is perpendicular to $mx+ny=18$,then

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