Question Number 10414 by amir last updated on 07/Feb/17
Commented by mrW1 last updated on 08/Feb/17
$${the}\:{angle}\:{between}\:{tangent}\:{lines}\:{is} \\ $$$${not}\:{a}\:{constant}! \\ $$
Commented by amir last updated on 08/Feb/17
$${hello}\:{dear}\:{mrW}\mathrm{1}.{thank}\:{you}\:{so}\:{much} \\ $$$${for}\:{solving}\:{this}\:{problems}. \\ $$$${yes}\:.{you}\:{are}\:{right}.\:{angle}\:{is}\:{not}\:{constant} \\ $$$${and}\:{i}\:{shuld}\:{fix}\:{it}. \\ $$
Answered by mrW1 last updated on 08/Feb/17
$${Q}\mathrm{1}: \\ $$$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${y}'=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${tangent}\:{line}\:{at}\:{point}\:\left({t},\frac{\mathrm{1}}{{t}}\right)\:{on}\:{curve} \\ $$$${is} \\ $$$${y}−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left({x}−{t}\right) \\ $$$${since}\:{point}\:\left({a},−{a}\right)\:{lies}\:{on}\:{tangent}\:{line} \\ $$$$−{a}−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left({a}−{t}\right) \\ $$$${at}^{\mathrm{2}} +\mathrm{2}{t}−{a}=\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}{a}^{\mathrm{2}} }}{\mathrm{2}{a}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}},\:{a}\neq\mathrm{0} \\ $$$${i}.{e}.\:{from}\:{point}\:{A}\left({a},−{a}\right)\:{with}\:{a}\neq\mathrm{0} \\ $$$${there}\:{are}\:\mathrm{2}\:{tangent}\:{lines}: \\ $$$${AD}\:{and}\:{AE} \\ $$$${D}\left({t}_{\mathrm{1}} ,\frac{\mathrm{1}}{{t}_{\mathrm{1}} }\right)\:{and}\:{E}\left({t}_{\mathrm{2}} ,\frac{\mathrm{1}}{{t}_{\mathrm{2}} }\right) \\ $$$${slope}\:{of}\:{AD}=\beta_{\mathrm{1}} \\ $$$${slope}\:{of}\:{AE}=\beta_{\mathrm{2}} \\ $$$${angle}\:{between}\:{tangent}\:{lines}=\beta=\beta_{\mathrm{2}} −\beta_{\mathrm{1}} \\ $$$$\mathrm{tan}\:\beta_{\mathrm{1}} =−\frac{\mathrm{1}}{{t}_{\mathrm{1}} ^{\mathrm{2}} }=−\frac{\mathrm{1}}{\left(\frac{−\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)^{\mathrm{2}} }=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\beta_{\mathrm{2}} =−\frac{\mathrm{1}}{{t}_{\mathrm{2}} ^{\mathrm{2}} }=−\frac{\mathrm{1}}{\left(\frac{−\mathrm{1}−\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)^{\mathrm{2}} }=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\beta=\frac{−\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}}{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}×\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}} \\ $$$$=\frac{{a}^{\mathrm{2}} \left(−\mathrm{2}−{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }+\mathrm{2}+{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)}{\left(\mathrm{2}+{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left(\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)+{a}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} \sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{\left(\mathrm{2}+{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)+{a}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} \sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{\mathrm{4}+\mathrm{4}{a}^{\mathrm{2}} +{a}^{\mathrm{4}} −\mathrm{4}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)+{a}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} }\right)\neq{constant} \\ $$$$ \\ $$$${Q}\mathrm{2}: \\ $$$${perpendicular}\:{line}\:{at}\:{point}\:\left({t},\frac{\mathrm{1}}{{t}}\right) \\ $$$${on}\:{curve}\:{xy}=\mathrm{1}\:{is} \\ $$$${y}−\frac{\mathrm{1}}{{t}}={t}^{\mathrm{2}} \left({x}−{t}\right) \\ $$$${since}\:{point}\:{A}\left({a},−{a}\right)\:{lies}\:{on}\:{the}\:{line} \\ $$$$−{a}−\frac{\mathrm{1}}{{t}}={t}^{\mathrm{2}} \left({a}−{t}\right) \\ $$$${t}^{\mathrm{4}} −{at}^{\mathrm{3}} −{at}−\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −{at}−\mathrm{1}\right)=\mathrm{0} \\ $$$${since}\:{t}^{\mathrm{2}} +\mathrm{1}\neq\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −{at}−\mathrm{1}=\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2}} =\frac{{a}\pm\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${i}.{e}.\:{from}\:{point}\:{A}\left({a},−{a}\right)\:{there}\:{are} \\ $$$$\mathrm{2}\:{perpendicular}\:{lines}: \\ $$$${AB}\:{and}\:{AC}\:{with} \\ $$$${B}\left({t}_{\mathrm{1}} ,\frac{\mathrm{1}}{{t}_{\mathrm{1}} }\right)\:{and}\:{C}\left({t}_{\mathrm{2}} ,\frac{\mathrm{1}}{{t}_{\mathrm{2}} }\right) \\ $$$${slope}\:{of}\:{AB}=\alpha_{\mathrm{1}} \\ $$$${slope}\:{of}\:{AC}=\alpha_{\mathrm{2}} \\ $$$${angle}\:{between}\:{them}\:=\alpha=\alpha_{\mathrm{2}} −\alpha_{\mathrm{1}} \\ $$$$\mathrm{tan}\:\alpha_{\mathrm{1}} ={t}_{\mathrm{1}} ^{\mathrm{2}} =\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha_{\mathrm{2}} ={t}_{\mathrm{2}} ^{\mathrm{2}} =\left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +\mathrm{2}−{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}×\frac{\left({a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)^{\mathrm{2}} −\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{16}+\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)^{\mathrm{2}} \left({a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}×\frac{−\mathrm{2}{a}×\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{16}+\left({a}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=−\frac{{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${angle}\:{between}\:{tangent}\:{line}\:{and} \\ $$$${perpendicular}\:{line}=\gamma=\beta_{\mathrm{1}} −\alpha_{\mathrm{1}} \\ $$$$\mathrm{tan}\:\gamma=\frac{−\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}−\frac{{a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}}{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}×\frac{{a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{2}{a}^{\mathrm{2}} +\left(\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left({a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}{\mathrm{2}\left(\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)+{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}\: \\ $$$$=−\frac{\mathrm{2}{a}^{\mathrm{2}} +\left(\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left({a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}{\mathrm{4}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)−\mathrm{4}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }+{a}^{\mathrm{4}} +{a}^{\mathrm{3}} \sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}\: \\ $$$$\Rightarrow\gamma=\mathrm{tan}^{−\mathrm{1}} \left(−\frac{\mathrm{2}{a}^{\mathrm{2}} +\left(\mathrm{2}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left({a}^{\mathrm{2}} +\mathrm{2}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}{\mathrm{4}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)−\mathrm{4}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }+{a}^{\mathrm{4}} +{a}^{\mathrm{3}} \sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}\right) \\ $$
Commented by mrW1 last updated on 08/Feb/17
Answered by mrW1 last updated on 08/Feb/17
$${Q}\mathrm{3}: \\ $$$${x}_{{B}} =\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${y}_{{B}} =\frac{\mathrm{1}}{{x}_{{B}} }=\frac{−{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${x}_{{D}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}} \\ $$$${y}_{{D}} =\frac{\mathrm{1}}{{x}_{{D}} }=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}} \\ $$$$ \\ $$$${area}\:{F}={F}_{\mathrm{1}} +{F}_{\mathrm{2}} −{F}_{\mathrm{3}} −{F}_{\mathrm{4}} \\ $$$${F}_{\mathrm{1}} ={a}\left({x}_{{B}} −{x}_{{D}} \right)={a}\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}−\frac{−\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)= \\ $$$$=\frac{{a}^{\mathrm{2}} +{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${F}_{\mathrm{2}} =\int_{{x}_{{D}} } ^{{x}_{{B}} } \frac{\mathrm{1}}{{x}}{dx}=\left(\mathrm{ln}\:{x}\right)\mid_{{x}_{{D}} } ^{{x}_{{B}} } =\mathrm{ln}\:\frac{{x}_{{B}} }{{x}_{{D}} }=\mathrm{ln}\:\frac{\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}}{\frac{−\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}} \\ $$$$=\mathrm{ln}\:\mid\frac{{a}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}{\mathrm{2}\left(\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }−\mathrm{1}\right)}\mid \\ $$$${F}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{B}} −{a}\right)\left({a}+{y}_{{B}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}−{a}\right)\left({a}+\frac{−{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)=\frac{\left(\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−{a}\right)\left(\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}+{a}\right)}{\mathrm{8}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{x}_{{D}} \right)\left({a}+{y}_{{D}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left({a}−\frac{−\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)\left({a}+\frac{\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)=\frac{\left({a}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left({a}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)}{\mathrm{2}{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}+{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${F}=\frac{{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{\mathrm{2}}+\mathrm{ln}\:\mid\frac{{a}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)}{\mathrm{2}\left(\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }−\mathrm{1}\right)}\mid \\ $$