The equation of the tangent to the curve $y=1-\mathrm{e}^{\frac{x}{3}}$ at the point of intersection with Y…
The equation of the tangent to the curve $y=1-\mathrm{e}^{\frac{x}{3}}$ at the point of intersection with Y -axis is
- $x-3 y=0$
- $x+3 y=0$
- $x+2 y=0$
- $3 x^{\prime}+y=0$
Solution
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
$\begin{aligned}
& y-0=\frac{-1}{3}(x-0) \\
& \Rightarrow 3 y=-x \\
& \Rightarrow x+3 y=0
\end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 2)
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