Question Number 40826 by behi83417@gmail.com last updated on 28/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18
$${A}\left({a},{a}\right)\:\:{B}\left(\alpha,\beta\right) \\ $$$${mid}\:{point}\:{AB}\left(\frac{{a}+\alpha}{\mathrm{2}},\frac{{a}+\beta}{\mathrm{2}}\right) \\ $$$${h}=\frac{{a}+\alpha}{\mathrm{2}}\:\:\:\:{k}=\frac{{a}+\beta}{\mathrm{2}} \\ $$$$\alpha=\mathrm{2}{h}−{a}\:\:\:\:\beta==\mathrm{2}{k}−{a} \\ $$$$ \\ $$$$\left(\mathrm{2}{h}−{a}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\mathrm{2}{k}−{a}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ={a}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$${hence}\:{locus}\:{is} \\ $$$$\left(\mathrm{2}{x}−{a}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\mathrm{2}{y}−{a}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ={a}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 28/Jul/18
$${thank}\:{you}\:{sir}\:{tanmay}.{but}\:{midpoint} \\ $$$${of}\:{AB},{not}\:{located}\:{on}\:{corve}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18
$${point}\left({h},{k}\right)\:\:{is}\:{the}\:{mid}\:{point}...{but}\:{i}\:{have}\:{not}\:{put} \\ $$$${them}\:{on}\:{the}\:{curve}... \\ $$$$\left(\alpha,\beta\right)\:{point}\:{lies}\:{on}\:{the}\:{curve}\:{that}\:{is}\:{point}\:{B} \\ $$$${so}\:{h}=\frac{{a}+\alpha}{\mathrm{2}}\:\:\:\alpha=\mathrm{2}{h}−{a}\:\:\:\beta=\mathrm{2}{k}−{a} \\ $$$${now}\:{i}\:{can}\:{put}\:{them}\:{in}\:{curve}\:... \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 28/Jul/18
$${ok}.{thanks}. \\ $$