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
Answer & explanation
Correct answer: option 1
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
$y-\frac{1}{e}= -\frac{e}{2}\left(x+\frac{1}{2}\right)$
$\Rightarrow 4 e y-4 =e(-2 e x-e) \Rightarrow 2 e(e x+2 y)+e^2-4=0$