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Question Number 171336 by cortano1 last updated on 13/Jun/22

Answered by mr W last updated on 13/Jun/22

$$k=\frac{y}{x}$$$$x^2-10x+k^2{x}^2-10kx+41=0$$$$(1+k^2)x^2-10(1+k)x+41=0$$$$\mathrm{\Delta }={10}^2{(1+k)}^2-4\times 41(1+k^2)⩾0$$$$25(1+2k)-16-16k^2$$$$8k^2-25k+8⩽0$$$$\Rightarrow \frac{25-3\sqrt{41}}{16}⩽k⩽\frac{25+3\sqrt{41}}{16}$$$$\Rightarrow k_{max}=\frac{25+3\sqrt{41}}{16}$$$$\Rightarrow k_{min}=\frac{25-3\sqrt{41}}{16}$$

Commented by cortano1 last updated on 13/Jun/22

$$yes$$

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