The equation of the normal to the curve ` y= e^(-2|x|)` at the point where the curve cuts the line ` x=-(1)/(2), ` is

Options

A. ` 2e(ex +2y)=4-e^(2) `
B. ` 2e(ex -2y)=e^(2)-4 `
C. ` 2e(ey -2x)=e^(2)-4 `
D. ` 2e(ey +2x)=e^(2)-4 `

Explanation

To find the equation of the normal to the curve \( y = e^{-2|x|} \) at the point where the curve intersects the line \( x = -\frac{1}{2} \), we can follow these steps:

### Step 1: Determine the point of intersection
Since we are interested in the point where \( x = -\frac{1}{2} \), we need to find the corresponding \( y \) value.

For \( x < 0 \), we have:
\[ y = e^{-2(-x)} = e^{2x} \]
Substituting \( x = -\frac{1}{2} \):
\[ y = e^{2 \cdot -\frac{1}{2}} = e^{-1} = \frac{1}{e} \]
Thus, the point of intersection is:
\[ \left(-\frac{1}{2}, \frac{1}{e}\right) \]

### Step 2: Find the slope of the tangent
To find the slope of the tangent line at this point, we need to differentiate \( y \) with respect to \( x \):
\[ y = e^{-2|x|} \quad \text{(for } x < 0 \text{, this becomes } y = e^{2x}) \]
Differentiating:
\[ \frac{dy}{dx} = 2e^{2x} \]
Now, substituting \( x = -\frac{1}{2} \):
\[ \frac{dy}{dx} = 2e^{2 \cdot -\frac{1}{2}} = 2e^{-1} = \frac{2}{e} \]
So, the slope of the tangent line at the point is \( \frac{2}{e} \).

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{2}{e}} = -\frac{e}{2} \]

### Step 4: Use the point-slope form to find the equation of the normal
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = \left(-\frac{1}{2}, \frac{1}{e}\right) \) and \( m = -\frac{e}{2} \):
\[ y - \frac{1}{e} = -\frac{e}{2}\left(x + \frac{1}{2}\right) \]

### Step 5: Simplify the equation
Expanding this:
\[ y - \frac{1}{e} = -\frac{e}{2}x - \frac{e}{4} \]
Rearranging gives:
\[ y = -\frac{e}{2}x - \frac{e}{4} + \frac{1}{e} \]
To combine the constants:
\[ y = -\frac{e}{2}x + \left(\frac{1}{e} - \frac{e}{4}\right) \]
Finding a common denominator for the constants:
\[ \frac{1}{e} - \frac{e}{4} = \frac{4 - e^2}{4e} \]
Thus, the equation becomes:
\[ y = -\frac{e}{2}x + \frac{4 - e^2}{4e} \]

### Step 6: Rearranging to standard form
Multiplying through by \( 4e \) to eliminate the fraction:
\[ 4ey = -2e^2x + (4 - e^2) \]
Rearranging gives:
\[ 2e^2x + 4ey + e^2 - 4 = 0 \]

### Final Equation
The final equation of the normal is:
\[ 2e^2x + 4ey + e^2 - 4 = 0 \]