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Question-171336




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

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