Find the equation of tangent and normal to the curve $y=2e^{-x/3}$ at the point where the curve meets the $y$-axis.

The given curve is $y=2e^{-x/3}......\left(1\right)$

given, curve (1) meets y-axis at $P(x, y)$ then $x=0$

$\Rightarrow y=2;\therefore P=(0,2)$

diff (1) w.r.t to $x,$ $\frac{dy}{dx}=2\cdot e^{\frac{-x}{3}}\cdot \left(\frac{-1}{3}\right)=\frac{-2}{3}{e}^{\frac{-x}{3}}$

Slope of the tangent at $P(0,2)$ is

${\left.m=\frac{dy}{dx}\right]}_{(0,2)}=\frac{-2}{3}$

equation of the tangent at $P(0,2)$ is

$y-y_1=m\left(x-x_1\right)$

$y-2=\frac{-2}{3}(x-0)\Rightarrow 2x+3y-6=0$

equation of the normal at $P(0,2)$ is

$y-y_1=\frac{-1}{m}\left(x-x_1\right)\Rightarrow y-2=\frac{3}{2}(x-0)$

$\Rightarrow 3x-2y+4=0$