Question Number 208755 by efronzo1 last updated on 22/Jun/24
Answered by mr W last updated on 22/Jun/24
Commented by mr W last updated on 22/Jun/24
$${y}=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}{x}=−\frac{\left({x}−\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{8} \\ $$$${y}=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{8}\:{in}\:{new}\:{system}\:\left({red}\right) \\ $$$${touching}\:{point}\:{at}\:\left({p},\:{q}\right) \\ $$$${q}=−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{8} \\ $$$$−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=\frac{{dy}}{{dx}}=−{p} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{{p}} \\ $$$${p}={r}\:\mathrm{cos}\:\theta=\frac{{rp}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}\:\Rightarrow{r}=\sqrt{\mathrm{1}+{p}^{\mathrm{2}} } \\ $$$${q}={r}+{r}\:\mathrm{sin}\:\theta={r}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}\right)=−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{8} \\ $$$$\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}\right)=−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{8} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }=−{p}^{\mathrm{2}} +\mathrm{14} \\ $$$${p}^{\mathrm{4}} −\mathrm{32}{p}^{\mathrm{2}} +\mathrm{192}=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{16}−\mathrm{8}=\mathrm{8} \\ $$$${r}=\sqrt{\mathrm{1}+\mathrm{8}}=\mathrm{3}\:\checkmark \\ $$
Commented by mr W last updated on 22/Jun/24
Commented by efronzo1 last updated on 23/Jun/24
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