# Tricky question to set up the DE for... probably easy to solve

• Aug 2nd 2010, 06:41 AM
HD09
Tricky question to set up the DE for... probably easy to solve
Find all curves in the first quadrant such that for any point, the segment of the tangent going from the y intercept to the x intercept is bisected by the point of tangency.

Here's my thoughts ...

Let $(x_0,y_0)$ be the point of tangency.

The tangent has the equation $y=\left.\frac{dy}{dx}\right\vert_{x_0}(x-x_0)+y_0$

The condition that the segment of the tangent going from the y-intercept to the x-intercept be bisected at the point of tangency means:

$y(0)=-\left.\frac{dy}{dx}\right\vert_{x_0}x_0+y_0=2y_0 \leftrightarrow y_0=-\left.\frac{dy}{dx}\right\vert_{x_0}x_0$

... or putting $y(2x_0)=0$ gives the same.

Now I'm stuck.

Thanks for any help!
• Aug 2nd 2010, 10:35 AM
Ackbeet
Well, it has to be true for every point on the graph of the function. Hence, can't you just use the DE

$-y(x)=y'(x)\,x$?