Question

The equation of tangent to the curve $y=1-e^{\frac{x}{2}}$ at the point where it meets y - axis is -

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

Answer 1

The correct option is A $x+2y=0$

Substituting $y=0$
$0=1-e^{\frac{x}{2}}\Rightarrow x=0$
Thus the point is $(0,0)$
Now, $\frac{dy}{dx}=-e^{\frac{x}{2}}\times \frac{1}{2}$
${\left(\frac{dy}{dx}\right)}_{(0,0)}=-\frac{1}{2}\times e^{\frac{0}{2}}=-\frac{1}{2}$
Therefore, the required equation is given by,
$y-0=\frac{-1}{2}(x-0)$
$2y=-x\Rightarrow x+2y=0$
Hence, option 'A' is correct.