Question Number 217146 by a.lgnaoui last updated on 02/Mar/25
$$\mathrm{determiner}\:\mathrm{le}\:\mathrm{cote}\:\mathrm{du}\:\mathrm{care}\:\boldsymbol{\mathrm{ABCD}} \\ $$$$\mathrm{inscrit}\:\mathrm{dans}\:\mathrm{l}\:\mathrm{elipse}\:\left\{\left(−\mathrm{3},+\mathrm{3}\right):\left(−\mathrm{8},+\mathrm{8}\right)\right\} \\ $$
Commented by a.lgnaoui last updated on 02/Mar/25
Answered by mr W last updated on 03/Mar/25
$${square}\:{s}×{s}\:{inscribed}\:{in}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${we}\:{have} \\ $$$$\frac{\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{s}=\frac{\mathrm{2}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${in}\:{this}\:{example}:\:{a}=\mathrm{3},\:{b}=\mathrm{8} \\ $$