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

**HD09** · August 2nd 2010, 05:41 AM

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!

**Ackbeet** · August 2nd 2010, 09:35 AM

Well, it has to be true for every point on the graph of the function. Hence, can't you just use the DE

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

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