Kim's question at Yahoo! Answers regarding finding point of tangency, given ODE and tangent line

MarkFL

Staff member
Here is the question:

How do you solve this AP calculus BC problem?

y'=(4x-2xy)/(x^2+y^2+1)

The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of the point P. Thank you!
I have posted a link there to this topic so the OP can see my work.

MarkFL

Staff member
Hello Kim,

First, we know the line through the origin, having slope $m=-1$ is given by:

$$\displaystyle y=-x$$

and we know at point $P(x,y)$, we must then have:

$$\displaystyle y'=-1$$

Using these two conditions, we then find, using the given ODE:

$$\displaystyle y'=\frac{4x+2x^2}{2x^2+1}=-1$$

Solving for $x$, we then obtain:

$$\displaystyle 4x+2x^2=-2x^2-1$$

$$\displaystyle 4x^2+4x+1=0$$

$$\displaystyle (2x+1)^2=0$$

$$\displaystyle x=-\frac{1}{2}\implies y=\frac{1}{2}$$

Hence:

$$\displaystyle P(x,y)=\left(-\frac{1}{2},\frac{1}{2} \right)$$

Here is a plot of the curve described by the ODE and the tangent line: