Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 98

Suppose that a tangent line is drawn to the hyperbola $xy = c$ at a point $P$.

a.) Prove that the midpoint of the line segment cut from this tangent line lay the coordinate axes is $P$.
b.) Prove that the triangle formed by the tangent line and the coordinate axes always has the same area, no
matter where $P$ is located on the hyperbola.


$
\begin{equation}
\begin{aligned}
\text{a.) since } xy &= c,\\
y &= \frac{c}{x}\\
\frac{dy}{dx} &= c \frac{d}{dx}\left(\frac{1}{x}\right)\\
\frac{dy}{dx} &= c \left(\frac{-1}{x^2}\right)\\
\frac{dy}{dx} &= -\frac{c}{x^2}
\end{aligned}
\end{equation}
$


Now we can get the tangent line through $\displaystyle P \left(x_1, \frac{c}{x}\right)$ by using point slope form.

$
\begin{equation}
\begin{aligned}
y -y_1 &= m(x-x_1)\\
y - \left(\frac{c}{x_1}\right) &= \frac{-c}{x_1^2}(x-x_1)\\
y-\frac{c}{x_1} &= \frac{-c}{x_1^2} x + \frac{c}{x_1}\\
y &= \frac{-c}{x_1^2} + \frac{2c}{x_1}
\end{aligned}
\end{equation}
$

Notice that the $y$-intercept $\displaystyle \frac{2c}{x_1}$ is twice the $y$-coordinate of $P$.

Solving for $x$-intercept,

$
\begin{equation}
\begin{aligned}
y &= \frac{c}{x_1^2} + \frac{2c}{x_1}\\
0 &= \frac{c}{x_1^2}x + \frac{2c}{x_1}\\
\frac{\cancel{c}x}{x_1^\cancel{2}} &= \frac{2\cancel{c}}{\cancel{x_1}}\\
x &= 2x_1
\end{aligned}
\end{equation}
$


It also shows that the $x$-intercept $2x_1$ is twice the $x$-coordinate of $P$. Therefore, the
midpoint of the line segment cut from the tangent line by the coordinate axes is $P$

b.) Solving for Area of triangle,


$
\begin{equation}
\begin{aligned}
\text{Area } &= \frac{1}{2}bh\\
\text{Area } &= \frac{1}{2} \quad \text{(x}\text{-intercept)} (y-\text{intercept)}\\
\text{Area } &= \frac{1}{\cancel{2}} \quad (\cancel{2}\cancel{x_1})\left(\frac{2c}{\cancel{x_1}}\right)\\
\text{Area } &= 2c
\end{aligned}
\end{equation}
$


It shows that no matter where $P$ is located on the hyperbola, the triangle formed by the tangent line and
coordinate axes always has the same area since the area is independent of point $P$.