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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

$$({at}_1^2,2{at}_1)({at}_2^2,2{at}_2)$$$$k=\frac{2{at}_2-2{at}_1}{{at}_2^2-{at}_1^2}=\frac{2a(t_2-t_1)}{a(t_2+t_1)(t_2-t_1)}=\frac{2}{t_2+t_1}$$$$\mathit{m}\mathit{i}\mathit{d}\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\alpha =\frac{{at}_1^2+{at}_2^2}{2}\beta =\frac{2{at}_2+2{at}_1}{2}$$$$\alpha =\frac{a}{2}(t_1^2+t_2^2)$$$$\beta =a(t_2+t_1)$$$$y=a\left(\frac{2}{k}\right)=\frac{2a}{k}$$$$$$$$$$$$$$

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

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

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

$$mostwelcome$$

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