The equation of the tangent to the curve \(y=1-\mathrm{e}^{\frac{x}{3}}\) at the point of intersection with Y -axis is

Given equation of curve is
\(y=1-\mathrm{e}^{\frac{x}{3}}...(i)\)
Since, curve intersects Y-axis, \(x=0\)
\(\begin{aligned}
& \therefore \quad y=1-\mathrm{e}^{\frac{0}{3}}=1-1 \\
& \Rightarrow y=0
\end{aligned}\)
\(\therefore \quad\) Tangent to the curve passes through origin
\(\therefore \quad\) Slope of tangent \(=\frac{\mathrm{d} y}{\mathrm{~d} x}\)
\(\therefore \quad\) Differentiating (i) w.r.to \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-\mathrm{e}^{\frac{x^3}{3}}}{3}\)
\(\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(0,0)}=\frac{-\mathrm{e}^{\frac{0}{3}}}{3}=\frac{-1}{3}\)
\(\therefore \quad\) Equation of tangent is
\(\begin{aligned}
& y-0=\frac{-1}{3}(x-0) \\
& \Rightarrow 3 y=-x \\
& \Rightarrow x+3 y=0
\end{aligned}\)