Question Number 65281 by mr W last updated on 27/Jul/19
Commented by mr W last updated on 27/Jul/19
$${segment}\:{of}\:{circle}\:{with}\:{size}\:\mathrm{2}{c}×{d}.\:\left({d}\leqslant{c}\right) \\ $$$${find}\:{the}\:{inscribed}\:{ellipse}\:{with} \\ $$$${maximum}\:{area}. \\ $$
Answered by mr W last updated on 28/Jul/19
$${R}={radius}\:{of}\:{circle} \\ $$$${d}\left(\mathrm{2}{R}−{d}\right)={c}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{\mathrm{2}{d}} \\ $$$${let}\:{h}={R}−{d}=\frac{{c}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}{d}} \\ $$$${eqn}.\:{of}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}+{h}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{hy}+{h}^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{hy}−{c}^{\mathrm{2}} =\mathrm{0}\:\:\:\:…\left({i}\right) \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{y}^{\mathrm{2}} −\frac{\mathrm{2}{a}^{\mathrm{2}} }{{b}}{y}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){y}^{\mathrm{2}} +\mathrm{2}\left({h}+\frac{{a}^{\mathrm{2}} }{{b}}\right){y}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${due}\:{to}\:{tangency}\:{there}\:{is}\:{only}\:{one}\:{root}, \\ $$$$\Delta=\mathrm{4}\left({h}+\frac{{a}^{\mathrm{2}} }{{b}}\right)^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\left({hb}+{a}^{\mathrm{2}} \right)^{\mathrm{2}} +{c}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{0}\:\:…\left({iii}\right) \\ $$$$\Rightarrow\left({hab}+{a}^{\mathrm{3}} \right)^{\mathrm{2}} +{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${let}\:{P}={ab} \\ $$$$\Rightarrow\left({hP}+{a}^{\mathrm{3}} \right)^{\mathrm{2}} +{c}^{\mathrm{2}} \left({P}^{\mathrm{2}} −{a}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${for}\:{maximum}\:{ellipse}\:{area},\:\frac{{dP}}{{da}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left({hP}+{a}^{\mathrm{3}} \right)\left(\mathrm{3}{a}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left(−\mathrm{4}{a}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}\left({hb}+{a}^{\mathrm{2}} \right)−\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${if}\:{h}=\mathrm{0},\:{i}.{e}.\:{d}={c}: \\ $$$$\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}=\frac{\sqrt{\mathrm{6}}{c}}{\mathrm{3}}\Rightarrow{b}=\frac{\sqrt{\mathrm{2}}{c}}{\mathrm{3}} \\ $$$${if}\:{h}\neq\mathrm{0}: \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{{h}}\left(\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{3}}−{a}^{\mathrm{2}} \right) \\ $$$${insert}\:{this}\:{into}\:\left({iii}\right): \\ $$$$\left(\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{3}}−{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} +{c}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{9}}+\frac{\mathrm{1}}{{h}^{\mathrm{2}} }\left(\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{3}}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{4}{c}^{\mathrm{2}} {h}^{\mathrm{2}} }{\mathrm{9}}+\left(\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{3}}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} −{h}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{4}} −\left(\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} \right){a}^{\mathrm{2}} +\frac{\mathrm{4}{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} −\sqrt{\left(\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{\mathrm{16}{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{9}}}\right] \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} −\sqrt{\frac{\mathrm{16}{c}^{\mathrm{4}} +\mathrm{24}{c}^{\mathrm{2}} {h}^{\mathrm{2}} +\mathrm{9}{h}^{\mathrm{4}} −\mathrm{16}{c}^{\mathrm{4}} −\mathrm{16}{c}^{\mathrm{2}} {h}^{\mathrm{2}} }{\mathrm{9}}}\right] \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} −{h}\sqrt{\frac{\mathrm{8}{c}^{\mathrm{2}} }{\mathrm{9}}+{h}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}{c}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} −{h}\sqrt{\frac{\mathrm{8}{c}^{\mathrm{2}} }{\mathrm{9}}+{h}^{\mathrm{2}} }\right)} \\ $$
Commented by mr W last updated on 28/Jul/19