Question Number 181759 by mr W last updated on 29/Nov/22 | ||
Commented by mr W last updated on 30/Nov/22 | ||
$${BC}={a} \\ $$$$\frac{{AB}}{{AC}}={k}={constant} \\ $$$${what}\:{is}\:{the}\:{locus}\:{of}\:{point}\:{A}? \\ $$ | ||
Answered by mr W last updated on 30/Nov/22 | ||
Commented by mr W last updated on 30/Nov/22 | ||
$${if}\:{k}=\mathrm{1},\:{the}\:{locus}\:{of}\:{A}\:{is}\:{the}\: \\ $$$${perpendicular}\:{bisector}\:{of}\:{BC}. \\ $$$${now}\:{we}\:{assume}\:{k}<\mathrm{1}. \\ $$$${c}^{\mathrm{2}} ={d}^{\mathrm{2}} +{e}^{\mathrm{2}} −\mathrm{2}{de}\:\mathrm{cos}\:\theta \\ $$$${b}^{\mathrm{2}} ={d}^{\mathrm{2}} +\left({a}+{e}\right)^{\mathrm{2}} −\mathrm{2}{d}\left({a}+{e}\right)\:\mathrm{cos}\:\theta \\ $$$$\left(\frac{{c}}{{b}}\right)^{\mathrm{2}} =\frac{{d}^{\mathrm{2}} +{e}^{\mathrm{2}} −\mathrm{2}{de}\:\mathrm{cos}\:\theta}{{d}^{\mathrm{2}} +\left({a}+{e}\right)^{\mathrm{2}} −\mathrm{2}{d}\left({a}+{e}\right)\:\mathrm{cos}\:\theta}={k}^{\mathrm{2}} \: \\ $$$${k}^{\mathrm{2}} {d}^{\mathrm{2}} +{k}^{\mathrm{2}} \left({a}+{e}\right)^{\mathrm{2}} −\mathrm{2}{k}^{\mathrm{2}} {d}\left({a}+{e}\right)\:\mathrm{cos}\:\theta={d}^{\mathrm{2}} +{e}^{\mathrm{2}} −\mathrm{2}{de}\:\mathrm{cos}\:\theta \\ $$$$\left(\mathrm{1}−{k}^{\mathrm{2}} \right){d}^{\mathrm{2}} +{e}^{\mathrm{2}} −{k}^{\mathrm{2}} \left({a}+{e}\right)^{\mathrm{2}} =\mathrm{2}{d}\left[\left(\mathrm{1}−{k}^{\mathrm{2}} \right){e}−{k}^{\mathrm{2}} {a}\right]\:\mathrm{cos}\:\theta \\ $$$${set}\:\left(\mathrm{1}−{k}^{\mathrm{2}} \right){e}−{k}^{\mathrm{2}} {a}=\mathrm{0},\:{i}.{e}.\:{e}=\frac{{k}^{\mathrm{2}} {a}}{\mathrm{1}−{k}^{\mathrm{2}} }=\frac{{a}}{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\left(\mathrm{1}−{k}^{\mathrm{2}} \right){d}^{\mathrm{2}} +{e}^{\mathrm{2}} −{k}^{\mathrm{2}} \left({a}+{e}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}−{k}^{\mathrm{2}} \right){d}^{\mathrm{2}} −\frac{{k}^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{1}−{k}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{d}=\frac{{ka}}{\mathrm{1}−{k}^{\mathrm{2}} }=\frac{{a}}{\frac{\mathrm{1}}{{k}}−{k}}={constant} \\ $$$${that}\:{means}\:{the}\:{locus}\:{of}\:{A}\:{is}\:{a}\:{circle} \\ $$$${with}\:{radius}\:\frac{{a}}{\frac{\mathrm{1}}{{k}}−{k}}\:{and}\:{center}\:{at}\:{point} \\ $$$${D}\:{with}\:{distance}\:\frac{{a}}{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{1}}\:{to}\:{point}\:{B}. \\ $$ | ||