Question Number 83262 by peter frank last updated on 29/Feb/20
Answered by mr W last updated on 29/Feb/20
$${curve}\:\mathrm{1}:\:{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{8}\:\:\:…\left({i}\right) \\ $$$${curve}\:\mathrm{2}:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} =\mathrm{4}\:\:\:…\left({ii}\right) \\ $$$${intersection}: \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{6}{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{y}=\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}} \\ $$$$\left({i}\right)+\left({ii}\right)×\mathrm{2}: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} =\mathrm{16}\:\Rightarrow{x}=\pm\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${tangent}\:{of}\:{curve}\:\mathrm{1}\:{at}\:{intersection}: \\ $$$$\mathrm{2}{x}+\mathrm{8}{yy}'=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{1}} ={y}'=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{x}}{{y}}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\right)=\mp\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${tangent}\:{of}\:{curve}\:\mathrm{2}\:{at}\:{intersection}: \\ $$$$\mathrm{2}{x}−\mathrm{4}{yy}'=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} ={y}'=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}}{{y}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\right)=\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{tan}\:\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right)=\frac{\pm\sqrt{\mathrm{2}}−\left(\mp\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}{\mathrm{1}+\left(\pm\sqrt{\mathrm{2}}\right)\left(\mp\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}=\pm\infty \\ $$$$\mid\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \mid=\frac{\pi}{\mathrm{2}} \\ $$
Commented by peter frank last updated on 29/Feb/20
$${thank}\:{you} \\ $$