Question Number 138470 by mathocean1 last updated on 13/Apr/21
$${Given}\:\left(\Gamma\right):\:{x}^{\mathrm{4}} −\mathrm{16}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${Show}\:{that}\:\left(\Gamma\right)\:{is}\:{a}\:{reunion}\:{of}\:\:{and}\: \\ $$$${Ellipsis}\:{and}\:{an}\:{hyperbole}\:{then}\:{give} \\ $$$${their}\:{equations}. \\ $$
Answered by MJS_new last updated on 14/Apr/21
$${x}^{\mathrm{4}} −\mathrm{16}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)\right)\left({x}^{\mathrm{2}} +\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)\right)=\mathrm{0}\:\vee\:\left({x}^{\mathrm{2}} +\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)=\mathrm{0}\:\Leftrightarrow\:{y}=\mathrm{1}\pm\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\mathrm{hyperbola} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)=\mathrm{0}\:\Leftrightarrow\:{y}=\mathrm{1}\pm\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}\:\mathrm{ellipse} \\ $$