Question Number 49737 by ajfour last updated on 09/Dec/18
Commented by ajfour last updated on 09/Dec/18
$${Find}\:{parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{of} \\ $$$${ellipse}\:{circumscribing}\:{a}\:{rectangle} \\ $$$${of}\:{sides}\:\boldsymbol{{l}}\:{and}\:\boldsymbol{{h}}. \\ $$
Answered by ajfour last updated on 09/Dec/18
$$\frac{{l}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{h}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{4}\:\:,\:\:{let}\:{T}\:=\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${T}\:=\:{a}^{\mathrm{2}} \left(\frac{{h}^{\mathrm{2}} }{\mathrm{4}−\frac{{l}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:=\:\frac{{a}^{\mathrm{4}} {h}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} −{l}^{\mathrm{2}} } \\ $$$$\frac{{dT}}{{da}}\:=\:\frac{\mathrm{4}{h}^{\mathrm{2}} {a}^{\mathrm{3}} \left(\mathrm{4}{a}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)−\mathrm{8}{a}^{\mathrm{5}} {h}^{\mathrm{2}} }{\left(\mathrm{4}{a}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{{dT}}{{da}}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\:{a}\:=\:\frac{{l}}{\:\sqrt{\mathrm{2}}}\:\:\:,\:\:{b}\:=\:\frac{{h}}{\:\sqrt{\mathrm{2}}}\:. \\ $$