The lines $3 x + y = 0$ may be perpendicular to one of the lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ .

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Solution

The line $3 x + y = 0$ may be perpendicular to one of the line given by $a x^{2} + 2 h x y + b y^{2} = 0$
Auxiliary equation $= b m^{2} + 2 h m + a = 0 ⟶ (1) 3 x + y = 0$
slope $(m_{1}) = - 3$
Hence slope of the line perpendicular to given line.
$∴$ Slope of one of the lines represented by the joint equation is $\frac{1}{3} .$
Now $\frac{1}{3}$ must be one of the roots of equation
$b {(\frac{1}{3})}^{2} + 2 h (\frac{1}{3}) + a = 0 b \frac{1}{9} + \frac{2 h}{3} + 0$
$b + 6 h + 9 a = 0$ (The required condition)

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