Question Number 43404 by peter frank last updated on 10/Sep/18
Answered by alex041103 last updated on 10/Sep/18
$$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\right) \\ $$$$\frac{{x}}{\mathrm{8}}−\frac{{y}}{\mathrm{2}}\left(\frac{{dy}}{{dx}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{x}}{\mathrm{4}{y}} \\ $$$${for}\:{x}=\mathrm{4}{sec}\:{t}\:{and}\:{y}=\mathrm{2}{tan}\:{t} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}{sec}\:{t}}{\mathrm{4}×\mathrm{2}\:{tan}\:{t}}=\frac{\mathrm{1}}{\mathrm{2}{sin}\:{t}} \\ $$$${the}\:{equation}\:{for}\:{the}\:{tangent}\:{line} \\ $$$${is}\:{y}−\mathrm{2}{tan}\:{t}=\frac{{dy}}{{dx}}\left({t}\right)\:\left({x}−\mathrm{4}{sec}\:{t}\right) \\ $$$$\Rightarrow{y}−\mathrm{2}\frac{{sin}\:{t}}{{cos}\:{t}}=\frac{\mathrm{1}}{\mathrm{2}{sin}\:{t}}\left({x}−\mathrm{4}{sec}\:{t}\right) \\ $$$$\mathrm{2}{ysin}\:{t}\:−\:\frac{\mathrm{4}}{{cos}\:{t}}{sin}^{\mathrm{2}} {t}={x}−\frac{\mathrm{4}}{{cos}\:{t}} \\ $$$$\mathrm{2}{ysin}\:{t}={x}−\frac{\mathrm{4}}{{cos}\:{t}}\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right) \\ $$$$\Rightarrow\mathrm{2}\:{y}\:{sin}\:{t}\:=\:{x}\:−\:\mathrm{4}\:{cos}\:{t} \\ $$
Commented by peter frank last updated on 10/Sep/18
$$ \\ $$$$ \\ $$$${thanks}\:{very}\:{much} \\ $$