Robin's question at Yahoo Answers regarding the osculating circle of a parabola

[MarkFL]

Here is the question:

How to find the center of an osculating circle?


osculating circles of the parabola y=1/2*x^2 at (-1,1/2)
I got the radius to be 2^(3/2) which is right.
All I need to know is how to find the center.

I have posted a link there to this thread so the Op can view my work.

 

[MarkFL]

Hello Robin,

For a curve in the plane expressed as a function $y(x)$ in Cartesian coordinates, the radius of curvature is given by:

\(\displaystyle R(x)=\left|\frac{\left(1+y'^2 \right)^{\frac{3}{2}}}{y''} \right|\)

We are given the curve:

\(\displaystyle y(x)=\frac{1}{2}x^2\)

Hence:

\(\displaystyle y'=x\)

\(\displaystyle y''=1\)

And so the radius of curvature for this function is given by:

\(\displaystyle R(x)=(1+x^2)^{\frac{3}{2}}\)

Hence:

\(\displaystyle R(-1)=2\sqrt{2}\)

Now, the center of the osculating circle will lie aline the normal line at the given point. The slope $m$ of this normal line is the negative multiplicative inverse of the slope of the tangent line. Thus:

\(\displaystyle m=-\frac{1}{\left.y'(-1) \right|_{x=-1}}=-\frac{1}{-1}=1\)

Thus, using the point-slope formula, the normal line is given by:

\(\displaystyle y-\frac{1}{2}=x+1\)

Now, the distance from the tangent point of the osculating circle and its center $\left(x_C,y_C \right)$ is the radius of curvature we found above, and so we may write:

\(\displaystyle \left(x_C+1 \right)^2+\left(y_C-\frac{1}{2} \right)^2=\left(2\sqrt{2} \right)^2\)

Since the center of the circle lies on the normal line we found, we have:

\(\displaystyle \left(x_C+1 \right)^2+\left(x_C+1 \right)^2=8\)

\(\displaystyle \left(x_C+1 \right)^2=4\)

\(\displaystyle x_C=-1\pm2\)

Since the curve is concave up, we take the root:

\(\displaystyle x_C=1\implies y_C=\frac{5}{2}\)

Thus, the center of the osculating circle at the given point is:

\(\displaystyle \left(x_C,y_C \right)=\left(1,\frac{5}{2} \right)\)