The equation of the normal to the curve $y=\cosh x$ drawn at the point nearest to the origin is

$\because y=\cos h x$ $\Rightarrow y^{\prime}=\sin \mathrm{hx}$ To find the nearest point to the origin, we can set the derivative equal to zero. $$ \therefore \sin \mathrm{h}(\mathrm{x})=0 $$ The only solution of this equation is $x=0$ At $x=0, y=\operatorname{cash}(0)=1$ The slope of normal is $$ \mathrm{m}=\left(-\frac{1}{\mathrm{y}^1}\right)_{(0,1)}=\left(-\frac{1}{\sinh x}\right)_{(0,1)} $$ $$ \mathrm{m}=\frac{-1}{0} $$ Now, the equation of the normal is $$ (y-1)=-\frac{1}{0}(x-0) \Rightarrow(x-0)=0 \Rightarrow x=0 $$