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June 20th, 2016, 09:24 AM   #1
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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.
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June 21st, 2016, 04:28 AM   #2
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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 y-axis 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 x-axis at $\displaystyle \left(x_0- \frac{y_0}{y'(x_0)}, 0\right)$.

The point mid-way 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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