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Question-45602




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

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