Single Variable Calculus, Chapter 7, 7.3-1, Section 7.3-1, Problem 66

Determine the equation of the tangent line to the curve $y = e^{-x}$ that is perpendicular to the line $2x - y = 8$

Since the tangent line is perpendicular to the tangent line, its slope is equal to the negative reciprocal of the perpendicular line. So..

$M_T = - \frac{1}{M_N}$

Solving for $M_N$


$
\begin{equation}
\begin{aligned}

& zx - y = 8
\\
& y = 2x - 8

\end{aligned}
\end{equation}
$


By observation, $M_N = 2$, hence, $\displaystyle M_T = -\frac{1}{2}$

Also, recall that the first derivative is equal to the slope of the tangent line at the curve, so..


$
\begin{equation}
\begin{aligned}

\text{if } y =& e^{-x}, \text{then}
\\
\\
y' =& e^{-x} (-1)

\end{aligned}
\end{equation}
$


Thus,


$
\begin{equation}
\begin{aligned}

\frac{-1}{2} =& -e^{-x}
\\
\\
\frac{1}{2} =& \frac{1}{e^x}
\\
\\
e^x =& 2

\end{aligned}
\end{equation}
$


Taking the natural logarithm of both sides..


$
\begin{equation}
\begin{aligned}

x (ln e) =& ln (2)
\\
\\
x(1) =& ln (2)
\\
\\
x =& ln (2)

\end{aligned}
\end{equation}
$


So when $x = ln(2)$, then...


$
\begin{equation}
\begin{aligned}

y =& e^{-x} = \frac{1}{e^x}
\\
\\
y =& \frac{1}{e^{e^{ln2}}} = \frac{1}{2}

\end{aligned}
\end{equation}
$


Therefore, by using point slope form, the equation of the tangent line is..


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - \frac{1}{2} =& \frac{-1}{2} (x - ln(2))
\\
\\
y =& \frac{-1}{2} x + \frac{ln (2)}{2} + \frac{1}{2}
\\
\\
y =& \frac{-x}{2} + \frac{ln(2) + 1}{2}



\end{aligned}
\end{equation}
$