MHB Solving Point of Tangency for AP Calculus BC Problem

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To solve the AP Calculus BC problem, the derivative of the curve is given by y'=(4x-2xy)/(x^2+y^2+1). The line through the origin with slope -1 is represented by y=-x, which leads to the condition y'=-1 at point P. By substituting this condition into the ODE, the equation simplifies to 4x+2x^2=-2x^2-1, resulting in the quadratic equation 4x^2+4x+1=0. Solving this yields x=-1/2, and consequently, y=1/2, giving the coordinates of point P as (-1/2, 1/2). The solution is confirmed with a plot of the curve and the tangent line.
MarkFL
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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.
 
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Hello Kim,

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

$$y=-x$$

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

$$y'=-1$$

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

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

Solving for $x$, we then obtain:

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

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

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

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

Hence:

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

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

View attachment 1430
 

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