The equation of the tangent to the curve y=1- | vedclass

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(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.Since the curve intersects the $Y$-axis,we set $x=0$.Substituting $x=0$ in $(i)$,we

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$x+3y=0$

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(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.Since the curve intersects the $Y$-axis,we set $x=0$.Substituting $x=0$ in $(i)$,we get $y=1-e^{0}=1-1=0$.Thus,the point of intersection is $(0, 0)$.The slope of the tangent is given by $\frac{dy}{dx}$.Differentiating $(i)$ with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{3} e^{\frac{x}{3}}$.At the point $(0, 0)$,the slope $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{3} e^{0} = -\frac{1}{3}$.The equation of the tangent line passing through $(0, 0)$ with slope $m = -\frac{1}{3}$ is $y - 0 = -\frac{1}{3}(x - 0)$.Multiplying by $3$,we get $3y = -x$,which simplifies to $x + 3y = 0$.

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

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