Question Number 46838 by peter frank last updated on 01/Nov/18 | ||
Answered by MrW3 last updated on 03/Nov/18 | ||
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }{y}'=\mathrm{0}\Rightarrow\frac{{x}}{{a}^{\mathrm{2}} }+\frac{{y}}{{b}^{\mathrm{2}} }{y}'=\mathrm{0} \\ $$$${e}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$ \\ $$$${eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−{c}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\left({say}\right) \\ $$$$\mathrm{2}\left({x}−{c}\right)+\mathrm{2}{yy}'=\mathrm{0}\Rightarrow\left({x}−{c}\right)+{yy}'=\mathrm{0} \\ $$$$ \\ $$$${contact}\:{points}:\:\left({ea},{k}\right)\:{and}\:\left({ea},−{k}\right) \\ $$$$\frac{{e}^{\mathrm{2}} {a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${e}^{\mathrm{2}} +\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{e}^{\mathrm{2}} {b}^{\mathrm{2}} +{k}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$\left({ea}−{c}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\frac{{ea}}{{a}^{\mathrm{2}} }+\frac{{k}}{{b}^{\mathrm{2}} }{m}=\mathrm{0} \\ $$$$\frac{{eb}^{\mathrm{2}} }{{a}}+{km}=\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$${ea}−{c}+{km}=\mathrm{0}\:\:\:...\left({iv}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${e}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({ea}−{c}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} −{r}^{\mathrm{2}} \:\:\:...\left({v}\right) \\ $$$$\left({iii}\right)−\left({iv}\right): \\ $$$$\frac{{eb}^{\mathrm{2}} }{{a}}−{ea}+{c}=\mathrm{0} \\ $$$$\Rightarrow{c}={e}\left({a}−\frac{{b}^{\mathrm{2}} }{{a}}\right)=\frac{{e}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}}={e}^{\mathrm{3}} {a} \\ $$$${from}\:\left({v}\right): \\ $$$${r}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)+\left({ea}−{c}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)+\frac{{e}^{\mathrm{2}} {b}^{\mathrm{4}} }{{a}^{\mathrm{2}} }={b}^{\mathrm{2}} \left[\mathrm{1}−\frac{{e}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right.}{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{4}} \right)={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)\left(\mathrm{1}−{e}^{\mathrm{4}} \right)={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} −{e}^{\mathrm{4}} +{e}^{\mathrm{6}} \right) \\ $$$$\Rightarrow{eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−{e}^{\mathrm{3}} {a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{4}} \right) \\ $$$${y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{ae}^{\mathrm{3}} {x}+{a}^{\mathrm{2}} {e}^{\mathrm{6}} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} −{e}^{\mathrm{4}} +{e}^{\mathrm{6}} \right) \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{ae}^{\mathrm{3}} {x}={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} −{e}^{\mathrm{4}} \right) \\ $$ | ||
Commented by peter frank last updated on 04/Nov/18 | ||
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mrW}_{\mathrm{3}} \\ $$ | ||