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Question-39709




Question Number 39709 by Raj Singh last updated on 10/Jul/18
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
what is the meaning of T^3 T′
$${what}\:{is}\:{the}\:{meaning}\:{of}\:{T}^{\mathrm{3}} {T}'\:\:\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
let point T is ct,(c/t)  eqn of normal  (y−(c/t))=(−(1/m))(x−ct)  x(dy/dx)+y=0   so (dy/dx)=((−y)/(x ))  hence  m=((−(c/t))/(ct))  m=((−1)/t^2 )     so  ((−1)/m)=t^2   so normal eqn  (y−(c/t))=(t^2 )(x−ct)  now let T′=(ct_1 ,(c/t_1 ))  ((c/t_1 )−(c/t))=t^2 (ct_1 −ct)  (((ct−ct_1 )/(tt_1 )))−t^2 (ct_1 −ct)=0 ihave done some mistske  edited    (ct−ct_1 )((1/(tt_1 ))+t^2 )=0  t^3 t_1 +1=0      so  t^3 t_1 =−1  point T(ct,(c/t))  and point T′((c/(−t^3 )),((−ct^3 )/))
$${let}\:{point}\:{T}\:{is}\:{ct},\frac{{c}}{{t}} \\ $$$${eqn}\:{of}\:{normal}\:\:\left({y}−\frac{{c}}{{t}}\right)=\left(−\frac{\mathrm{1}}{{m}}\right)\left({x}−{ct}\right) \\ $$$${x}\frac{{dy}}{{dx}}+{y}=\mathrm{0}\:\:\:{so}\:\frac{{dy}}{{dx}}=\frac{−{y}}{{x}\:}\:\:{hence}\:\:{m}=\frac{−\frac{{c}}{{t}}}{{ct}} \\ $$$${m}=\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }\:\:\:\:\:{so}\:\:\frac{−\mathrm{1}}{{m}}={t}^{\mathrm{2}} \\ $$$${so}\:{normal}\:{eqn}\:\:\left({y}−\frac{{c}}{{t}}\right)=\left({t}^{\mathrm{2}} \right)\left({x}−{ct}\right) \\ $$$${now}\:{let}\:{T}'=\left({ct}_{\mathrm{1}} ,\frac{{c}}{{t}_{\mathrm{1}} }\right) \\ $$$$\left(\frac{{c}}{{t}_{\mathrm{1}} }−\frac{{c}}{{t}}\right)={t}^{\mathrm{2}} \left({ct}_{\mathrm{1}} −{ct}\right) \\ $$$$\left(\frac{{ct}−{ct}_{\mathrm{1}} }{{tt}_{\mathrm{1}} }\right)−{t}^{\mathrm{2}} \left({ct}_{\mathrm{1}} −{ct}\right)=\mathrm{0}\:{ihave}\:{done}\:{some}\:{mistske} \\ $$$${edited} \\ $$$$ \\ $$$$\left({ct}−{ct}_{\mathrm{1}} \right)\left(\frac{\mathrm{1}}{{tt}_{\mathrm{1}} }+{t}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${t}^{\mathrm{3}} {t}_{\mathrm{1}} +\mathrm{1}=\mathrm{0}\:\:\:\:\:\:{so}\:\:{t}^{\mathrm{3}} {t}_{\mathrm{1}} =−\mathrm{1} \\ $$$${point}\:{T}\left({ct},\frac{{c}}{{t}}\right)\:\:{and}\:{point}\:{T}'\left(\frac{{c}}{−{t}^{\mathrm{3}} },\frac{−{ct}^{\mathrm{3}} }{}\right) \\ $$
Commented by Raj Singh last updated on 10/Jul/18
how you find (ct,c/t)
$${how}\:{you}\:{find}\:\left({ct},{c}/{t}\right) \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
ct,(c/t)  is a point lies on the curve...parsmetric
$${ct},\frac{{c}}{{t}}\:\:{is}\:{a}\:{point}\:{lies}\:{on}\:{the}\:{curve}…{parsmetric} \\ $$

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