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Find all plane curves in the first quadrant such that for every point $P$ on the curve, $P$ bisects the part of the tangent line at $P$ that lies in the first quadrant?

To my understanding this problem requires deriving and solving of a differential equation, however I am stuck in the deriving part. Could somebody help me understand how to derive the differential equation for this one?

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up vote 5 down vote accepted

Since the tangent line to any graph $y = f(x)$ at $(a,b)$ goes through $(a,f(a))$ and has slope $f'(a)$, it has equation $$ y - f(a) = (x - a) f'(a) $$ This line intersects the $x$-axis at the point $(a - f(a)/f'(a), 0)$ and the y-axis at $(0, f(a) - a f'(a))$. The midpoint of the segment determined by these two points is $(\frac{1}{2} (a - f(a)/f'(a)), \frac{1}{2} (f(a) - a f'(a)))$. We are trying to find the functions $f$ for which this is $(a, f(a))$. Equating the first components of these things we get $$ a = \frac{1}{2} (a - f(a)/f'(a)) $$ and equating the second components we get $f(a) = \frac{1}{2} (f(a) - a f'(a))$, which turns out to convey exactly the same information as the first equation (after rewriting it a little).

Changing $a$ to the more familiar variable $x$ and letting $y = f(x)$ the relation above is $x = \frac{1}{2} (x - y/y')$ which implies $2x = x - y/y'$ or that $x = -y/y'$ or in what is perhaps a more familiar form $$ y' = -y/x. $$ This is a separable differential equation and can be solved with the usual methods; the general solution is $y = C/x$ with $C$ a constant--- but in your case $C$ is not quite arbitrary: we must have $C > 0$ if this curve is to have a portion lying in the first quadrant.

You will probably have to do a little bit of work to fully justify why this discussion finds all curves meeting the conditions, and not just some of them, but this is probably the basic idea.

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