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(D) Given the homogeneous equation $ax^{2}+2hxy+by^{2}=0 \quad \dots(i)$This represents a pair of straight lines passing through the origin. Let the s

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$h^{2}-ab$

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(D) Given the homogeneous equation $ax^{2}+2hxy+by^{2}=0 \quad \dots(i)$This represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m$ and $m_{1}$.Thus,the lines are $y=mx$ and $y=m_{1}x$.Their product is $(y-mx)(y-m_{1}x) = 0$,which simplifies to $y^{2}-(m+m_{1})xy+mm_{1}x^{2}=0$.Dividing the original equation by $b$,we get $y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0 \quad \dots(ii)$.Comparing $(i)$ and $(ii)$,we have:$m+m_{1} = -\frac{2h}{b}$ and $mm_{1} = \frac{a}{b}$.Since $m$ is a root of the quadratic equation $bm^{2}+2hm+a=0$ (obtained by substituting $y=mx$ in the original equation):$bm^{2}+2hm+a=0$$bm^{2}+hm = -a-hm$$m(bm+h) = -(a+hm)$This does not lead directly to the target. Let us use the relation $m_{1} = -\frac{2h}{b}-m$.Substituting into $mm_{1} = \frac{a}{b}$:$m(-\frac{2h}{b}-m) = \frac{a}{b}$$-2hm - bm^{2} = a$$bm^{2} + 2hm + a = 0$Adding $h^{2}$ to both sides:$bm^{2} + 2hm + h^{2} = h^{2} - a$This is not quite right. Let's re-evaluate: $bm^{2}+2hm+a=0 \implies bm^{2}+hm = -a-hm$. Actually,from $bm^{2}+2hm+a=0$,we have $bm^{2}+hm = -a-hm$. Wait,the expression is $(h+bm)^{2} = h^{2} + 2hbm + b^{2}m^{2}$.Since $bm^{2}+2hm+a=0$,then $b^{2}m^{2}+2hbm+ab=0$.Thus $b^{2}m^{2}+2hbm = -ab$.Adding $h^{2}$ to both sides: $h^{2}+2hbm+b^{2}m^{2} = h^{2}-ab$.Therefore,$(h+bm)^{2} = h^{2}-ab$.

If $m$ is the slope of one of the lines represented by $ax^{2}+2hxy+by^{2}=0$,then $(h+bm)^{2}$ is equal to

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