Solving Point of Tangency for AP Calculus BC Problem

[MarkFL]

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]

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:

View attachment 1430