Answer is: option2
\( x + 2y - 4 = 0 \)Solution:
\[ \begin{cases} x = e^t \\ y = 2e^{-t} \end{cases} \]
Solution I. When \( t = \ln 2 \), we have the point \( \left(e^{\ln 2}, 2e^{-\ln 2}\right) = (2,1) \).
The slope of the curve is \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2e^{-t}}{e^t}. \]
When \( t = \ln 2 \), \[ \frac{dy}{dx} = \frac{-2e^{-\ln 2}}{e^{\ln 2}} = -\frac{1}{2}. \]
Then the equation of the desired tangent is: \[ y - 1 = -\frac{1}{2}(x - 2) \] \[ 2y - 2 = -x + 2 \] \[ x + 2y - 4 = 0. \]
Solution II. We eliminate the parameter to obtain \( y = \frac{2}{x} \). When \( t = \ln 2 \), \( x = 2 \), so \( y = 1 \).
Then \[ \frac{dy}{dx} = -\frac{2}{x^2}, \] so when \( x = 2 \), \[ \frac{dy}{dx} = -\frac{1}{2}. \] The equation is obtained as before.
Solution III. First we obtain the point desired point \( (2,1) \) as before. Then we note that only proposed answers (B) and (E) contain this point.
The slope of equation (B) is \( m = -\frac{1}{2} \). The slope of equation (E) is \( m = -2 \).
Then graph the given curve (either with the non-parametric equation \( y = \frac{2}{x} \) or with the given parametric equations) with a window that has equal units. It is clear that the slope of the curve at \( (2,1) \) is much closer to \( -\frac{1}{2} \) than to \( 2 \).
