Question

The equation of the tangent to the curve $y=1-e^{x/2}$ at the point of intersection with the y-axis is:

• A. x + 2y = 0
• B. 2x + y = 0
• C. x - y = 0
• D. none of these

Answer 1

Answer: (A)

Consider the equation of the curve;
$y=1-e^{\frac{x}{2}}$
Differentiate both side w.r.t. 'x'
$\frac{dy}{dx}=-\frac{1}{2}{e}^{\frac{x}{2}}$
The point of intersection of the curve with y-axis is x = 0, y = 0
$\therefore \frac{dy}{dx}{|}_{(0,0)}=\frac{-1}{2}$
$\therefore$ Equation of the tangent to the curve is given by
$(y-0)=\frac{dy}{dx}(x-0)$
$\Rightarrow y=\frac{-1}{2}(x)$
$\Rightarrow 2y+x=0$