Question

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

• A. -3
• B. -2
• C. 2
• D. 3

Answer 1

Answer: (D)

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