Question Number 225240 by Spillover last updated on 18/Oct/25
Answered by mr W last updated on 19/Oct/25
$$\left({a}\right) \\ $$$${R}=\frac{\mid\mathrm{3}×\mathrm{5}−\mathrm{4}×\mathrm{4}+\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$$\left({b}\right) \\ $$$${y}=\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$${e}^{{x}} −{e}^{−{x}} −\mathrm{2}{y}=\mathrm{0} \\ $$$$\left({e}^{{x}} \right)^{\mathrm{2}} −\mathrm{2}{y}\left({e}^{{x}} \right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{x}} ={y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:\:\left(−\:{rejected}\right) \\ $$$$\Rightarrow{x}=\mathrm{ln}\:\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{y}=\mathrm{sinh}^{−\mathrm{1}} \:{x}=\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$
Commented by Spillover last updated on 25/Nov/25
$${thanks} \\ $$