Consider any tangent line drawn to the curve $y= \frac{k}{x}$ where $k$ is a constant. If $A$ and $B$ are the x-intercept and y-intercept of the tangent line and $O$ is the origin, prove that the area of the triangle $AOB$ is the same for all tangent lines to the curve.
I got as far as taking the derivative, writing the formula for the angle of the triangle, equating the slope of the line with the derivative, and that was it.
Any pointers or proof sketches?
Let the tangent touch $y=k/x$ at $x=a$ (so $y=k/a$). The equation of the tangent will be $$ y-\frac{k}{a} = m(a) (x-a), $$ where $m(a)$ is the value of the derivative at $x=a$. We have $$ \left. \frac{dy}{dx} \right|_{x=a} = \left. -\frac{k}{x^2} \right|_{x=a} = -\frac{k}{a^2}. $$
Therefore the tangent is given by $$ y-\frac{k}{a} = -\frac{k}{a^2} (x-a). $$ Now, the intercepts. The $x$-intercept occurs when $y=0$, so $$ 0 -\frac{k}{a} = -\frac{k}{a^2} (A-a) \implies A = 2a $$ On the other hand, the $y$-intercept has $x=0$, so $$ B-\frac{k}{a} = \frac{k}{a} \implies B = \frac{2k}{a}. $$ $\triangle AOB$ is right-angled, so its area is given by $$ \frac{1}{2}AB = \frac{1}{2} 2a \frac{2k}{a} = 2k, $$ independent of $a$.
The formula for the tangent line at a point $a, f(a)$ on the curve $y=f(x)$ is $$y=f\left(n\right)-\frac{k}{n^2}\left(x-n\right)$$ The $x$-intercept will be $$0=f\left(n\right)-\frac{k}{n^2}\left(x-n\right) \implies x=\frac{n^2}{k}f\left(n\right)+n$$ The $y$-intercept will be $$y=f\left(n\right)-\frac{k}{n^2}\left(0-n\right)\implies y=f\left(n\right)+\frac{k}{n}$$ Finding the area of the triangle isn't hard now; the vertical leg is the $y$-intercept and the horizontal leg is the $x$-intercept. We thus find that the area of the triangle is $$\frac{1}{2}\left(f\left(n\right)+\frac{k}{n}\right)\left(\frac{n^2}{k}f\left(n\right)+n\right)= \frac{(n f(n) + k)^2}{2 k}= \frac{(k + k)^2}{2 k}=\frac{1}{2k}$$ Where we used the fact that $f(n) = \frac{k}{n}$