Question Number 45608 by peter frank last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
$${p}\left(\alpha,\beta\right)\:\:{Q}\left({a},{b}\right)\:\:\:{pointR}\:{devides}\:{PQ}\:\mathrm{2}:\mathrm{1} \\ $$$${R}\left(\frac{\mathrm{2}{a}+\alpha}{\mathrm{3}},\frac{\mathrm{2}{b}+\beta}{\mathrm{3}}\right)=\left({h},{k}\right)\:{say} \\ $$$$\frac{\mathrm{2}{a}+\alpha}{\mathrm{3}}={h}\:\:\:\alpha=\mathrm{3}{h}−\mathrm{2}{a}\:\:\:\: \\ $$$$\frac{\mathrm{2}{b}+\beta}{\mathrm{3}}={k}\:\:\:\:\:\beta=\mathrm{3}{k}−\mathrm{2}{b} \\ $$$${since}\:\alpha,\beta\:\:{lies}\:{o}\:{n}\:{hyperbola}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${so}\:\frac{\alpha^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\beta^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${caution}\:{here}\:{co}−{ordinate}\:{of}\:{point}\:{Q}\:{is}\:{given} \\ $$$$\left({a},{b}\right)\:{wheras}\:{in}\:{hyperbola}\:{eqn}\:{already}\:\:{a},{b} \\ $$$${exists}...{so}\:{pls}\:{check}\:{whether}\:{they}\:{are}\:{identical} \\ $$$$... \\ $$$$\frac{\left(\mathrm{3}{h}−\mathrm{2}{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\left(\mathrm{3}{k}−\mathrm{2}{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${so}\:{the}\:{required}\:{locus}\:{is} \\ $$$$\frac{\left(\mathrm{3}{x}−\mathrm{2}{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\left(\mathrm{3}{y}−\mathrm{2}{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\:{pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 14/Oct/18
$$\mathrm{thank}\:\mathrm{sir}\:\mathrm{your}\:\mathrm{absolute}\:\mathrm{right} \\ $$