If \((m+3 n)(3 m+n)=4 h^2\), then the acute angle between the lines represented by \(m x^2+2 h x y+n y^2=0\) is

We have \(3\text m^2+10\text{mn}+3\text n^2=4\text h^2\qquad\ldots(1)\)
Given lines are \(m x^2+2 h x+n y^2=0\)
Let \(\mathrm{m}_1\) and \(\mathrm{m}_2\) be the slopes of the lines
\(\therefore \mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{n}} \text { and } \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{m}}{\mathrm{n}} \)
\( \text { Now }\left(\mathrm{m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2 \)
\( =\left(\frac{-2 \mathrm{~h}}{\mathrm{n}}\right)^2-\frac{4 \mathrm{~m}}{\mathrm{n}}=\frac{4 \mathrm{~h}^2-4 \mathrm{mn}}{\mathrm{n}^2}=\) \(\frac{3 \mathrm{~m}^2+6 \mathrm{mn}+3 \mathrm{n}^2}{\mathrm{n}^2} \ldots[\text { From (1)] } \)
\( =\frac{3(\mathrm{~m}+\mathrm{n})^2}{\mathrm{n}^2} \)
\( \therefore \mathrm{m}_1-\mathrm{m}_2=\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}}\)
\(\therefore\) Let \(\theta\) be the required angle.
\(\text {Now } \tan \theta=\frac{\left|\mathrm{m}_1-\mathrm{m}_2\right|}{1+\mathrm{m}_1 \mathrm{~m}_2} \)
\( =\frac{\left(\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}}\right)}{\left(1+\frac{\mathrm{m}}{\mathrm{n}}\right)}=\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}} \times \frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\sqrt{3} \Rightarrow \theta\) \(=60^{\circ}=\frac{\pi^{\mathrm{c}}}{3}\)