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  4. If equation of tangent to the curve y=-e(-x/2) where it crosses the y-axis is (x/a)+(y/b)=1, then (a-b) is equal to

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Q. If equation of tangent to the curve $y=-e^{\frac{-x}{2}}$ where it crosses the $y$-axis is $\frac{x}{a}+\frac{y}{b}=1$, then $(a-b)$ is equal to

Application of Derivatives

Solution:

At $ x=0, y=-1$,
Also, $ y^{\prime}(0)=\frac{1}{2}$
$\therefore $ Equation of tangent is $( y +1)=\frac{1}{2}( x -0)$
$\Rightarrow \frac{x}{2}+\frac{y}{-1}=1 \Rightarrow(a-b)=2-(-1)=3$.