To find the curve for which the part of tangent cut-off by the axes (the portion of the tangent between the coordinate axes) is bisected at the point of tangency.
Let $\dfrac{x}{a} + \dfrac{y}{b} = 1$ be the tangent. It cuts the axes at $(a,0)$ and $(0,b)$. So the mid point of the part of tangent cut-off by the axes is $(\dfrac{a}{2},\dfrac{b}{2})$. The slope of this tangent is $\dfrac{-b}{a}$. $(Since\ y = \dfrac{-b}{a}x + b).$
Let the slope of the required curve at point $(x,y)$ given by $\dfrac{dy}{dx} = f(x,y)$.
So we can say that $f(\dfrac{a}{2},\dfrac{b}{2}) = \dfrac{-b}{a} \Rightarrow f(\dfrac{a}{2},\dfrac{b}{2}) = \dfrac{-b/2}{a/2} \Rightarrow f(x,y) = \dfrac{-y}{x}$.
Now $f(x,y) = \dfrac{dy}{dx} = \dfrac{-y}{x} \Rightarrow \dfrac{dy}{y} = -\dfrac{dx}{x} \Rightarrow \log(y) = -\log(x) + \log(c) \Rightarrow xy = c$.
Is this the correct way to proceed? Any other ideas?
Best Answer
Another approach :
Let $\quad Y=Y(X)\quad$ the equation of the curve.
At point $(X,Y)$ the equation of the tangent is $\quad y=Y+(x-X)Y'(X)$
The intersects to axes are : $\begin{cases} 0=Y+(a-X)Y'(X) \quad\to\quad a=X-\frac{Y}{Y'(X)}\\ b=Y+(0-X)Y'(X) \quad\to\quad b=Y-XY'(X) \end{cases}$
Condition : $\quad \begin{cases} \frac{a}{2}=X \quad\to\quad 2X=X-\frac{Y}{Y'(X)} \\ \frac{b}{2}=Y \quad\to\quad 2Y=Y-XY'(X) \end{cases}\quad\to\quad \frac{Y'}{Y}+\frac{1}{X}=0$ $$Y(X)=\frac{C}{X}$$ The curve is a rectangular hyperbola (orthogonal asymptotes).
Note : This property of rectangular hyperbola is known for long (Apollonius theorem) http://mathworld.wolfram.com/Hyperbola.html