If (m+3n)(3m+n)=4h^2,then the acute angle bet | vedclass

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(A) Given the equation of the pair of lines: $mx^2+2hxy+ny^2=0$.The condition given is $(m+3n)(3m+n)=4h^2$.Expanding the condition: $3m^2+mn+9mn+3n^2=

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$\frac{\pi}{3}$

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(A) Given the equation of the pair of lines: $mx^2+2hxy+ny^2=0$.The condition given is $(m+3n)(3m+n)=4h^2$.Expanding the condition: $3m^2+mn+9mn+3n^2=4h^2 \Rightarrow 3m^2+10mn+3n^2=4h^2$.The formula for the tangent of the angle $	heta$ between the lines $ax^2+2hxy+by^2=0$ is $	an 	heta = \left|\frac{2\sqrt{h^2-ab}}{a+b}ight|$.Here,$a=m$ and $b=n$.Substituting the values: $	an 	heta = \left|\frac{2\sqrt{h^2-mn}}{m+n}ight|$.From the given condition,$4h^2 = 3m^2+10mn+3n^2$,so $h^2 = \frac{3m^2+10mn+3n^2}{4}$.Then $h^2-mn = \frac{3m^2+10mn+3n^2-4mn}{4} = \frac{3m^2+6mn+3n^2}{4} = \frac{3(m+n)^2}{4}$.Therefore,$\sqrt{h^2-mn} = \frac{\sqrt{3}|m+n|}{2}$.Substituting this into the tangent formula: $	an 	heta = \left|\frac{2 \cdot \frac{\sqrt{3}|m+n|}{2}}{m+n}ight| = \sqrt{3}$.Thus,$	heta = 60^{\circ} = \frac{\pi}{3}$.

If $(m+3n)(3m+n)=4h^2$,then the acute angle between the lines represented by $mx^2+2hxy+ny^2=0$ is

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