A curve is defined parametrically by $\mathrm{x}={\mathrm{e}}^{\sqrt{\mathrm{t}}}, \mathrm{y}=3\mathrm{t}-{\log }_{\mathrm{e}}\left({\mathrm{t}}^2\right),$ where t is a parameter. Then the equation of the tangent line drawn to the curve at t = 1 is

$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{\mathrm{e}}^{\sqrt{\mathrm{t}}}}{2\sqrt{\mathrm{t}}} \mathrm{and} \frac{\mathrm{dy}}{\mathrm{dt}}=3-\frac{2}{\mathrm{t}}.\\ \mathrm{At} \mathrm{t}=1, \mathrm{x}=\mathrm{e} \mathrm{and} \mathrm{y}=3\\ \therefore \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{e}}{2}, \frac{\mathrm{dy}}{\mathrm{dt}}=1.\\ \mathrm{So}, \mathrm{the} \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{is} \frac{2}{\mathrm{e}}.\\ \mathrm{Hence}, \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{is} \mathrm{y}=\frac{2}{\mathrm{e}}\mathrm{x}+1.\end{array}$