The pair of lines \(l x^2+2(l+m) x y+m y^2=0\) lies along two diameters of a circle and divides the circle…

1. \(\frac{1}{2}\)
2. \(\frac{2}{\sqrt{3}}\)
3. \(\frac{11}{12}\)
4. \(\frac{13}{12}\)

Solution

Let angle between given pair of lines is ' \(\theta\). According to given information, \(A_2=5 A_1 \Rightarrow \frac{1}{2}(\pi-\theta) r^2=\left(\frac{1}{2} \theta r^2\right) \times 5\)
\(\begin{array}{ll} \Rightarrow \pi-\theta=5 \theta \Rightarrow \theta=\frac{\pi}{6} \\ \cos \frac{\pi}{6}=\frac{l+m}{\sqrt{(l-m)^2+4(l+m)^2}} \\ \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{l+m}{\sqrt{5 l^2+5 m^2+6 l m}} \\ \Rightarrow \quad \frac{3}{4}=\frac{(l+m)^2}{5 l^2+5 m^2+6 l m} \\ \Rightarrow \quad 15 l^2+15 m^2+18 l m=4 l^2+4 m^2+8 l m \\ \Rightarrow \quad 11 l^2+11 m^2+10 l m=0 \\ \Rightarrow \quad 11\left(l^2+m^2+2 l m\right)-22 l m+10 l m=0 \\ \Rightarrow \quad 11(l+m)^2-12 l m=0 \\ \Rightarrow \quad \frac{l m}{(l+m)^2}=\frac{11}{12} \end{array}\) Hence, option (c) is correct.

Asked in: AP EAMCET 2019 (23 Apr Shift 1)

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