Question Number 171336 by cortano1 last updated on 13/Jun/22
Answered by mr W last updated on 13/Jun/22
$${k}=\frac{{y}}{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}{x}+{k}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{10}{kx}+\mathrm{41}=\mathrm{0} \\ $$$$\left(\mathrm{1}+{k}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{10}\left(\mathrm{1}+{k}\right){x}+\mathrm{41}=\mathrm{0} \\ $$$$\Delta=\mathrm{10}^{\mathrm{2}} \left(\mathrm{1}+{k}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{41}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\mathrm{25}\left(\mathrm{1}+\mathrm{2}{k}\right)−\mathrm{16}−\mathrm{16}{k}^{\mathrm{2}} \\ $$$$\mathrm{8}{k}^{\mathrm{2}} −\mathrm{25}{k}+\mathrm{8}\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{25}−\mathrm{3}\sqrt{\mathrm{41}}}{\mathrm{16}}\leqslant{k}\leqslant\frac{\mathrm{25}+\mathrm{3}\sqrt{\mathrm{41}}}{\mathrm{16}} \\ $$$$\Rightarrow{k}_{{max}} =\frac{\mathrm{25}+\mathrm{3}\sqrt{\mathrm{41}}}{\mathrm{16}} \\ $$$$\Rightarrow{k}_{{min}} =\frac{\mathrm{25}−\mathrm{3}\sqrt{\mathrm{41}}}{\mathrm{16}} \\ $$
Commented by cortano1 last updated on 13/Jun/22
$$\:{yes} \\ $$