Question

The equation of the normal to the curve $y=e^{-2|x|}$ at the point where the curve cuts the line $x=\frac{1}{2}$ is

• A. $2e(ex+2y)=e^2-4$
• B. $2e(ex-2y)=e^2-4$
• C. $2e(ey-2x)=e^2-4$
• D. none of these

Answer 1

The correct option is B $2e(ex-2y)=e^2-4$

Given equation of curve is
$y=e^{-2\left|x\right|}$
At $x=\frac{1}{2}\Rightarrow y=e^{-1}$
$\frac{dy}{dx}=-2e^{-2x}$
${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{x=\frac{1}{2}}=-2e^{-1}$
Slope of tangent at $(\frac{1}{2},e^{-1})=\frac{-2}{e}$
Slope of normal $=\frac{e}{2}$
Equation of normal at $(\frac{1}{2},e^{-1})$ is
$y-\frac{1}{e}=\frac{e}{2}(x-\frac{1}{2})$
$y-\frac{ex}{2}=\frac{1}{e}-\frac{e}{4}$
$\frac{2y-ex}{2}=\frac{4-e^2}{4e}$
$\Rightarrow 2e(ex-2y)=e^2-4$