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June 20th, 2016, 09:24 AM  #1 
Newbie Joined: Jun 2016 From: sri lanka Posts: 3 Thanks: 0  geometrical problem to solve using differential equation.. How to get the equation for this using differential equations... Find the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact.... Last edited by skipjack; June 20th, 2016 at 09:29 AM. 
June 21st, 2016, 04:28 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Have you not even started? Call the function y(x). The tangent line to the graph y= y(x) at point $\displaystyle (x_0, y_0)$ is $\displaystyle y= y'(x_0)(x x_0)+ y_0$. When x= 0, $\displaystyle y= y'(x_0)(0 x_0)+ y_0= y_0 x_0y'(x_0)$ so the tangent line will intersect the yaxis at $\displaystyle (0, y_0 x_0y'(x_0))$. When y= 0, $\displaystyle 0= y'(x_0)(x x_0)+ y_0$, $\displaystyle y'(x_0)(x x_0)= y_0$, $\displaystyle x x_0= \frac{y_0}{y'(x_0)}$, and $\displaystyle x= x_0 \frac{y_0}{y'(x_0)}$. So the tangent line intersects the xaxis at $\displaystyle \left(x_0 \frac{y_0}{y'(x_0)}, 0\right)$. The point midway between the two intersections is $\displaystyle \left(\frac{x_0}{2} \frac{y_0}{2y'(x_0)}, \frac{y_0}{2} \frac{x_0y'(x_0)}{2}\right)$. We are told that the point of tangency, $\displaystyle (x_0, y_0)$ is that midpoint so we must have $\displaystyle x_0= \frac{x_0}{2} \frac{y_0}{2y'(x_0)}$ or $\displaystyle x_0= \frac{y_0}{y'(x_0)}$ and $\displaystyle y_0= \frac{y_0}{2} \frac{y'(x_0)}{2}$ or $\displaystyle y_0= y'(x_0)$. Since that is to be true at any point of tangency, we must have $\displaystyle x= \frac{y}{y'(x)}$ or $\displaystyle xy'(x)= y$ and $\displaystyle y= y'(x)$. 

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