Question Number 45602 by peter frank last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
$$\left({ct},\frac{{c}}{{t}}\right)\:{lies}\:{on}\:{rectangulsr}\:{hyperbola} \\ $$$${xy}={c}^{\mathrm{2}} \\ $$$$\frac{{d}}{{dx}}\left({xy}\right)=\frac{{d}}{{dx}}\left({c}^{\mathrm{2}} \right) \\ $$$${x}\frac{{dy}}{{dx}}+{y}=\mathrm{0}\:\:\:\frac{{dy}}{{dx}}=−\frac{{y}}{{x}} \\ $$$${slope}\:{of}\:{tsngent}\:{at}\left({ct},\frac{{c}}{{t}}\right) \\ $$$${m}=\left(\frac{{dy}}{{dx}}\right)_{\left({ct},\frac{{c}}{{t}}\right)} =\frac{−\frac{{c}}{{t}}}{{ct}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${m}×{m}'=−\mathrm{1} \\ $$$$−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }×{m}'=−\mathrm{1}\:\:\:\:{m}'={t}^{\mathrm{2}} \\ $$$${normal}\:{eqn} \\ $$$${y}−\frac{{c}}{{t}}={t}^{\mathrm{2}} \left({x}−{ct}\right) \\ $$$${pls}\:{wait}… \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 17/Oct/18
$$\mathrm{am}\:\mathrm{still}\:\mathrm{waiting}\:\mathrm{sir}…. \\ $$
Answered by ajfour last updated on 14/Oct/18
$${xy}={c}^{\mathrm{2}} \\ $$$${A}\:{point}\:{on}\:{the}\:{hyperbola}\:\left({ct},\:\frac{{c}}{{t}}\right) \\ $$$$−\frac{{dx}}{{dy}}\:=\:{t}^{\mathrm{2}} \\ $$$${let}\:{point}\:{from}\:{where}\:{the}\:{normals} \\ $$$${are}\:{drawn}\:{be}\:\left({h},{k}\right) \\ $$$${eq}.\:{of}\:{normals}: \\ $$$${y}−{k}\:=\:{t}^{\mathrm{2}} \left({x}−{h}\right) \\ $$$${since}\:\left({ct},\:\frac{{c}}{{t}}\right)\:{lies}\:{on}\:{such}\:{normal}, \\ $$$${so}\:\:\:\:\:\:\:\frac{{c}}{{t}}−{k}\:=\:{t}^{\mathrm{2}} \left({ct}−{h}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{{ct}}^{\mathrm{4}} −\boldsymbol{{ht}}^{\mathrm{3}} +\boldsymbol{{kt}}−\boldsymbol{{c}}\:=\:\mathrm{0} \\ $$$$……. \\ $$