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Question

The lines $$3x+y=0$$ may be perpendicular to one of the lines given by $$ax^2+2hxy+by^2=0$$.


Solution

The line $$ 3x+y=0$$ may be perpendicular to one of the line given by $$ a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$$
Auxiliary equation$$=b{ m }^{ 2 }+2hm+a=0\longrightarrow (1)\\ 3x+y=0$$
 slope$$({ m }_{ 1 })=-3$$
 Hence slope  of the line perpendicular to given line.
$$ \therefore$$ Slope of one of the lines represented by the joint  equation is $$\cfrac { 1 }{ 3 } .$$
 Now $$\cfrac { 1 }{ 3 }$$ must be one of the roots of equation
$$ b{ \left( \cfrac { 1 }{ 3 }  \right)  }^{ 2 }+2h\left( \cfrac { 1 }{ 3 }  \right) +a=0\\ b\cfrac { 1 }{ 9 } +\cfrac { 2h }{ 3 } +0$$
$$ b+6h+9a=0\quad $$(The required condition)

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