Question Number 196723 by mr W last updated on 30/Aug/23
$${prove}\:{that}\:{the}\:{curve}\: \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4}\: \\ $$$${is}\:{an}\:{ellipse}\:{and}\:{find}\:{its}\:{semi} \\ $$$${major}\:{axis}\:{and}\:{semi}\:{minor}\:{axis}. \\ $$
Answered by Frix last updated on 30/Aug/23
$$\mathrm{Squaring},\:\mathrm{transforming}\:\mathrm{etc}.\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{y}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{1} \\ $$$$\mathrm{Testing}\:\mathrm{with}\:{y}^{\mathrm{2}} =\frac{\mathrm{12}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{gives} \\ $$$$\mid{x}+\mathrm{4}\mid+\mid{x}−\mathrm{4}\mid=\mathrm{8};\:\mathrm{true}\:\mathrm{for}\:−\mathrm{4}\leqslant{x}\leqslant\mathrm{4}\: \\ $$
Commented by mr W last updated on 30/Aug/23
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Answered by mr W last updated on 30/Aug/23
$${v}=\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${u}=\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${u}+{v}=\mathrm{4} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${u}^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{4}{x} \\ $$$$\Rightarrow{u}−{v}={x} \\ $$$$\Rightarrow{u}=\frac{\mathrm{4}+{x}}{\mathrm{2}} \\ $$$$\Rightarrow{v}=\frac{\mathrm{4}−{x}}{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{16}+{x}^{\mathrm{2}} \right)}{\mathrm{4}}=\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{12} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{ellipse}\:{with}\:{a}=\mathrm{2},\:{b}=\sqrt{\mathrm{3}} \\ $$