Practicing Success
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 |
$2 e(e x+2 y)=4-e^2$ $2 e(e x-2 y)=e^2-4$ $2 e(e y-2 x)=e^2-4$ $2 e(e y+2 x)=e^2-4$ |
$2 e(e x+2 y)=4-e^2$ |
The line $x=-\frac{1}{2}$ cuts the curve $y=e^{-2|x|}$ at $y=e^{-1}$. Thus, the coordinates of the point of intersection are $P(-1 / 2,1 / e)$. In the neighbourhood of $x=-\frac{1}{2}$, we have $y=e^{2 x}$ as the equation of the curve. ∴ $\frac{d y}{d x}=2 e^{2 x} \Rightarrow\left(\frac{d y}{d x}\right)_{x=-\frac{1}{2}}=\frac{2}{e}$ Equation of the normal at $P(-1 / 2,1 / e)$ is $y-\frac{1}{e}= -\frac{e}{2}\left(x+\frac{1}{2}\right)$ $\Rightarrow 4 e y-4 =e(-2 e x-e) \Rightarrow 2 e(e x+2 y)+e^2-4=0$ |