Question Number 50820 by ajfour last updated on 20/Dec/18
Commented by ajfour last updated on 20/Dec/18
$${Choose}\:{your}\:{origin}\:{and}\:{find}\: \\ $$$${equation}\:{of}\:{parabola}\:{such}\:{that} \\ $$$${it}\:{has}\:{maximum}\:{length}\:{inside} \\ $$$${ellipse}\:\left({parameters}\:{a}\:{and}\:{b}\right). \\ $$
Answered by ajfour last updated on 21/Dec/18
$${y}={Ax}^{\mathrm{2}} −{b} \\ $$$${b}\mathrm{sin}\:\theta\:=\:{Aa}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta−{b} \\ $$$${let}\:\:\mathrm{sin}\:\theta\:=\:{t} \\ $$$$\Rightarrow\:\:{Aa}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)−{b}\:=\:{bt} \\ $$$${other}\:{than}\:{t}=−\mathrm{1} \\ $$$$\:\:\:\:\mathrm{1}−{t}\:=\:\frac{{b}}{{Aa}^{\mathrm{2}} }\:\:\:\Rightarrow\:\:\:{t}=\mathrm{1}−\frac{{b}}{{Aa}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{b}\mathrm{sin}\:\theta\:=\:{y}_{{A}} \:=\:{b}−\frac{{b}^{\mathrm{2}} }{{Aa}^{\mathrm{2}} } \\ $$$$\:\:\:{x}\:=\:\sqrt{\frac{{b}+{y}}{{A}}}\:\:\:\Rightarrow\:\:\frac{{dx}}{{dy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{A}\left({b}+{y}\right)}} \\ $$$$\:\:\:{L}=\int_{{A}} ^{\:\:{C}} {dl}\:=\:\mathrm{2}\int_{−{b}} ^{\:\:{y}_{{A}} } \sqrt{\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} }\:{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\int_{−{b}} ^{\:\:{y}_{{A}} } \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{A}\left({b}+{y}\right)}}\:{dy} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\frac{{dL}}{{dA}}\:=\:\int_{−{b}} ^{\:\:{y}_{{A}} } \frac{\left(−\frac{\mathrm{1}}{\mathrm{4}{A}^{\mathrm{2}} \left({b}+{y}\right)}\right)}{\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{A}\left({b}+{y}\right)}}}\:{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{b}^{\mathrm{2}} }{{A}^{\mathrm{2}} {a}^{\mathrm{2}} }\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{A}\left(\mathrm{2}{b}−\frac{{b}^{\mathrm{2}} }{{Aa}^{\mathrm{2}} }\right)}} \\ $$$$\:\:{let}\:{I}=\:\int_{−{b}} ^{\:\:{y}_{{A}} } \frac{\frac{\mathrm{1}}{\left({b}+{y}\right)}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{A}\left({b}+{y}\right)}}}\:{dy} \\ $$$$\:\:\:\:\:\:=\int\frac{{dz}}{\:\sqrt{{z}^{\mathrm{2}} +\frac{{z}}{\mathrm{4}{A}}}}\:=\:\int\frac{{dz}}{\:\sqrt{\left({z}+\frac{\mathrm{1}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}{A}^{\mathrm{2}} }}} \\ $$$$=\:\:\mathrm{ln}\:\mid{z}+{c}+\sqrt{\left({z}+{c}\right)^{\mathrm{2}} −{c}^{\mathrm{2}} }\mid+{k} \\ $$$$\:\:\:\:\:\:\:\:{where}\:\left[\:\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}{A}}\:\:\:\right] \\ $$$${y}_{{A}} \:=\:{b}−\frac{{b}^{\mathrm{2}} }{{Aa}^{\mathrm{2}} }\:\:;\:\:{z}_{{A}} ={b}+{y}_{{A}} =\mathrm{2}{b}−\frac{\mathrm{2}{b}^{\mathrm{2}} {c}}{{a}^{\mathrm{2}} } \\ $$$${y}_{{B}} \:=\:−{b}\:\:\Rightarrow\:\:{z}_{{B}} \:=\:\mathrm{0} \\ $$$${let}\:\:{z}_{{A}} +{c}\:=\:{t}_{{A}} \:=\:\mathrm{2}{b}+\frac{\mathrm{1}}{\mathrm{2}{A}}\left(\mathrm{1}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:{t}_{{B}} \:=\:{z}_{{B}} +{c}\:=\frac{\mathrm{1}}{\mathrm{2}{A}} \\ $$$${I}\:=\:\mathrm{ln}\:\frac{{t}_{{A}} +\sqrt{{t}_{{A}} ^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{t}_{{B}} } \\ $$$$\frac{{dL}}{{dA}}\:=\:\mathrm{0}\:\:\:\Rightarrow \\ $$$$\frac{{I}}{\mathrm{8}{A}^{\mathrm{2}} }\:=\:\frac{{b}^{\mathrm{2}} }{{A}^{\mathrm{2}} {a}^{\mathrm{2}} }\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{A}\left(\mathrm{2}{b}−\frac{{b}^{\mathrm{2}} }{{Aa}^{\mathrm{2}} }\right)}} \\ $$$$\:\:\:\: \\ $$$$\Rightarrow\:\mathrm{ln}\:\frac{{t}_{{A}} +\sqrt{{t}_{{A}} ^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{t}_{{B}} }\:=\:\frac{\mathrm{8}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{At}_{{A}} }} \\ $$$$\Rightarrow\:{In}\:{here}\:\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}{A}} \\ $$$${with}\:{t}_{{A}} =\:\mathrm{2}{b}+\frac{\mathrm{1}}{\mathrm{2}{A}}\left(\mathrm{1}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\:\:;\:{t}_{{B}} =\frac{\mathrm{1}}{\mathrm{2}{A}} \\ $$$${A}\:{is}\:{obtainable}\:{from}\:{above}\:{equation}. \\ $$$${If}\:\:{for}\:{example}\:\:{A}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\sqrt{\left(\mathrm{1}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{8}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\sqrt{\mathrm{1}+\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\:\:\:{b}\:=\:\mathrm{0}\:\:\:\left({naturally}\:{if}\:{within}\:{an}\right. \\ $$$${ellipse}\:{with}\:{b}\rightarrow\mathrm{0}\:,\:{parabola}\:{shall} \\ $$$${have}\:{maximum}\:{length}\:{if}\:{A}=\mathrm{0}\:. \\ $$
Commented by ajfour last updated on 21/Dec/18
$${mrW}\:{Sir}\:{can}\:{you}\:{cast}\:{a}\:{glance} \\ $$$${please}\:.. \\ $$
Commented by mr W last updated on 21/Dec/18
$${please}\:{check}\:{what}\:{is}\:{c}\:{in} \\ $$$$\Rightarrow\:\mathrm{ln}\:\frac{{t}_{{A}} +\sqrt{{t}_{{A}} ^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{t}_{{B}} }\:=\:\frac{\mathrm{8}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{At}_{{A}} }} \\ $$$${in}\:{this}\:{eqn}.\:{we}\:{should}\:{only}\:{have}\:{a},{b},\:{A}. \\ $$
Commented by ajfour last updated on 21/Dec/18
$${Thank}\:{you}\:{Sir},\:\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}{A}}\:. \\ $$