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Question Number 119892 by peter frank last updated on 27/Oct/20

Answered by TANMAY PANACEA last updated on 28/Oct/20

(at_1 ^2 ,2at_1 ) (at_2 ^2 ,2at_2 ) k=((2at_2 −2at_1 )/(at_2 ^2 −at_1 ^2 ))=((2a(t_2 −t_1 ))/(a(t_2 +t_1 )(t_2 −t_1 )))=(2/(t_2 +t_1 )) mid point α=((at_1 ^2 +at_2 ^2 )/2) β=((2at_2 +2at_1 )/2) α=(a/2)(t_1 ^2 +t_2 ^2 ) β=a(t_2 +t_1 ) y=a((2/k))=((2a)/k)

$$\left({at}_{\mathrm{1}} ^{\mathrm{2}} ,\mathrm{2}{at}_{\mathrm{1}} \right)\:\left({at}_{\mathrm{2}} ^{\mathrm{2}} ,\mathrm{2}{at}_{\mathrm{2}} \right) \\ $$$${k}=\frac{\mathrm{2}{at}_{\mathrm{2}} −\mathrm{2}{at}_{\mathrm{1}} }{{at}_{\mathrm{2}} ^{\mathrm{2}} −{at}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{\mathrm{2}{a}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}{{a}\left({t}_{\mathrm{2}} +{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}=\frac{\mathrm{2}}{{t}_{\mathrm{2}} +{t}_{\mathrm{1}} } \\ $$$$\boldsymbol{{mid}}\:\boldsymbol{{point}}\:\alpha=\frac{{at}_{\mathrm{1}} ^{\mathrm{2}} +{at}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}\:\:\:\beta=\frac{\mathrm{2}{at}_{\mathrm{2}} +\mathrm{2}{at}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\alpha=\frac{{a}}{\mathrm{2}}\left({t}_{\mathrm{1}} ^{\mathrm{2}} +{t}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$$\beta={a}\left({t}_{\mathrm{2}} +{t}_{\mathrm{1}} \right) \\ $$$${y}={a}\left(\frac{\mathrm{2}}{{k}}\right)=\frac{\mathrm{2}{a}}{{k}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 28/Oct/20

thank you.

$$\mathrm{thank}\:\mathrm{you}. \\ $$

Commented by TANMAY PANACEA last updated on 29/Oct/20

most welcome

$${most}\:{welcome} \\ $$

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