Question Number 211180 by Durganand last updated on 30/Aug/24
Answered by mr W last updated on 30/Aug/24
$$\left({ax}+{by}+{c}\right)\left(\frac{{x}}{{a}}+\frac{{y}}{{b}}+{d}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right){xy}+\left({ad}+\frac{{c}}{{a}}\right){x}+\left({bd}+\frac{{c}}{{b}}\right)+{cd} \\ $$$$={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{sin}\:\theta}{xy} \\ $$$$\Rightarrow{cd}=\mathrm{0} \\ $$$$\Rightarrow{ad}+\frac{{c}}{{a}}=\mathrm{0} \\ $$$$\Rightarrow{bd}+\frac{{c}}{{b}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}}{{b}}+\frac{{b}}{{a}}=−\frac{\mathrm{2}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{c}={d}=\mathrm{0} \\ $$$${say}\:−\frac{{a}}{{b}}={k}=\mathrm{tan}\:\alpha_{\mathrm{1}} \\ $$$$\Rightarrow{k}+\frac{\mathrm{1}}{{k}}=\frac{\mathrm{2}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{k}^{\mathrm{2}} −\frac{\mathrm{2}{k}}{\mathrm{sin}\:\theta}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}\pm\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:{or}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\alpha_{\mathrm{1}} =\frac{\theta}{\mathrm{2}},\:\alpha_{\mathrm{2}} =\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\alpha_{\mathrm{1}} =\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}},\:\alpha_{\mathrm{2}} =\frac{\theta}{\mathrm{2}} \\ $$$${line}\:\mathrm{1}:\:{ax}+{by}=\mathrm{0}\:\Rightarrow{y}=−\frac{{a}}{{b}}{x}={kx}=\left(\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){x} \\ $$$${line}\:\mathrm{2}:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{0}\:\Rightarrow{y}=−\frac{{b}}{{a}}{x}=\frac{{x}}{{k}}=\frac{{x}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${angle}\:{between}\:{them}\:=\frac{\pi}{\mathrm{2}}\pm\theta\:\checkmark \\ $$
Commented by mr W last updated on 30/Aug/24
