Osculating circle

Question

Compute the radius of osculating circle of the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ at one of its vertices and give a geometrical method for the construction of this osculating circle. Consider the particular case of equilateral hyperbola $x^2-y^2=a^2$

I computed the radius on the vertex $(a,0)$ by finding the reciprocal of the curvature of hyperbola. Which is $r=\frac{1}{|\kappa|}=\frac {b^2}{a}. \ $

Now how can I solve the next part? "give a geometrical method for the construction of this osculating circle"

Answer 1

In the particular case of the equilateral hyperbola $$ x^{2} - y^{2} = a^{2}, $$ the center of the osculating circle is $$ \left(a + \frac{b^{2}}{a}, 0\right) = (2a, 0), $$ and the radius is $a$. Let $O$ be the origin (where the hyperbola's axes (and asymptotes) cross) and $V$ the vertex $(a, 0)$. Draw a circle of center $V$ through $O$, and let $P$ be the point diametrically opposite this circle on the axis. The osculating circle has center $P$ and passes through $V$.

Answer 2

If $\alpha : I \rightarrow \mathbb{R}^2, \alpha (t)=(x(t),y(t))$ is a regular parametrization of hyperbola, then the center of osculating circle at a point t is given by $$C (t) =\alpha (t) + \frac{1}{\kappa}N(t)$$,where $\kappa$ is the curvature of $\alpha$, and $N$ is unit normal vector field of $\alpha$.

Osculating circle approximates the curve in the neighborhood of a point, and you want the circle and the curve to have the same tangent vector and normal vector and, of course, the same curvature.

You have the center and radius of a circle, I think that's enough.