Question

Equation of tangent at that point of the curve $y=1-e^{\frac{x}{2}}$, where it meets y-axis

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

Answer 1

Answer: (A)

$x+2y=0$

Given ,

$y=1-e^{\frac{x}{2}}$ ……(1)

Differentiate w.r. t x both side we get

$\frac{dy}{dx}=-e^{\frac{x}{2}}\times \frac{1}{2}$ …..(2)

Given that graph ofequation $\left(1\right)$ meets y-axis

$\Rightarrow x=0$

Put in equation $\left(1\right),$ we get

$y=1-e^{\frac{0}{2}}$

$y=1-e^0$

$y=1-1$

$y=0$

Put $y=0$ in equation $\left(1\right)$ we get

$x=0$

Now, put the value of $x,y$ in equation $\left(2\right),$ we get

${\left(\frac{dy}{dx}\right)}_{(0,0)}=-e^{\frac{0}{2}}\times \frac{1}{2}=-1\times \frac{1}{2}=-\frac{1}{2}$

Now required equation is

$y-y_1=\frac{dy}{dx}(x-x_1)$

$y-0=-\frac{1}{2}(x-0)$

$y=-\frac{1}{2}x$

$2y+x=0$

This is the required equation of tangent.