Hyperbolic local replacement in the neighborhood of a curve's vertex The proof of the Rotation Index Theorem in Introduction to Riemannian Manifolds by John M. Lee, claims (without further elaboration) that, given a vertex $o$ on an otherwise smooth parametric curve and a ball $B(o,\epsilon)$, there is a hyperbola (see the dashed arc)



passing through the intersection points $p$ and $q$ of the curve with the circle, whose tangents at $p$ and $q$ equal the tangents to the curve, and such that the arc of the hyperbola between $p$ and $q$ lies inside the ball.
The angle between the tangents at $p$ and $q$ can be assumed to be in the open interval $(-\pi,\pi)$. We can also assume that these tangents meet at a point $v\in B(o,\epsilon)$.
A more precise statement of the desired property would be the following
Lemma. Consider two points $p$ and $q$ in the plane and two straight lines passing through them that intersect at a point $v$ not aligned with $p$ and $q$. If $p$, $q$ and $v$ belong in the ball $B(o,\epsilon)$, then there exists an arc of hyperbola passing through $p$ and $q$ such that

*

*both lines are tangent to the hyperbola at $p$ and $q$ and

*the arc of hyperbola between $p$ and $q$ lies inside the ball $B(o,\epsilon)$.
I've been thinking in the lemma with no luck, and would appreciate any useful hint on how to prove it.
Note: Even though we might also assume that $\|p-o\|=\|q-o\|=\epsilon$, I haven't included this hypothesis in the lemma because it looks irrelevant.
 A: Here's a geometrical construction of the hyperbola. It relies on a well-known property of conic sections: the line through the midpoint of two tangency points and through the intersection of the tangents, also passes through the centre of the conic.
In our case, the centre $O$ of the conic lies on line $MV$, where $M$ is the midpoint of $PQ$. The conic intersects segment $MV$ at some point $R$ which we are free to choose at will. To complete the construction, we can use another useful property: the tangent at $R$ is parallel to $PQ$ and is thus determined once $R$ is fixed.
This tangent intersects $PV$ at $U$: if $L$ is the midpoint of $PR$, line $LU$ passes through the centre $O$, which can then be constructed as the intersection of $LU$ and $MV$. Note that $O$ lies inside convex angle $\angle PVQ$ if $VR>RM$ (in this case the conic is an ellipse) and lies outside it if $VR<RM$ (in this case the conic is a hyperbola). The limiting case $VR=RM$ leads to $LU\parallel MV$, in which case the conic is a parabola. For a hyperbola, choose then $R$ such that $VR<RM$.
To complete the construction various methods can be used. One can, for instance, reflect $Q$ and $R$ about $O$ to get two more points $Q'$ and $R'$. The unique conic through $PQRQ'R'$ is then the solution.
Finally, the arc of conic section limited by chord $PQ$, being a convex figure, completely lies inside triangle $PQV$ and thus inside $B(o,\epsilon)$.

A: For the sake of completeness, here is a proof of the well-known property mentioned in the accepted answer
Theorem
Let $P$ and $Q$ be two points in a conic with equation
$$
    \mathbf x^TA\mathbf x+2\mathbf b^T\mathbf x+c=0
$$
and $M=(P+Q)/2$ its mid point. If $V$ is the intersection of the tangents to the conic at $P$ and $Q$ and the conic is not a parabola, then the center of the conic lies on the line passing through $M$ and $V$.
Proof. Without loss of generality we may assume that $V=(0,0)$. First recall that the tangent to the conic at a point $Z$ is
$$
\mathbf x^TAZ + \mathbf b^T\mathbf x+\mathbf b^TZ+c=0,
$$
which, under our assumptions, leads to
$$
\mathbf b^TP + c = 0,\quad P^TAP=c\quad
 \textrm{ and}\quad \mathbf b^TQ + c = 0,\quad Q^TAQ=c \tag1
$$
and
$$
\mathbf x^TAP + \mathbf b^T\mathbf x = 0
    \quad\textrm{ and}\quad
\mathbf x^TAQ + \mathbf b^Tx = 0.
$$
We have to show that there exists $\lambda\in\mathbb R$ for which the equation of the conic becomes
$$
\big(\mathbf x^T-\frac\lambda2(P+Q)^T\big)A
\big(\mathbf x-\frac\lambda2(P+Q)\big)+d=0
$$
for some constant $d$. It is clear that this happens if, and only if, the equation becomes
$$
\mathbf x^TA\mathbf x-\lambda(P+Q)^TA\mathbf x
+\frac{\lambda^2}4(P+Q)^TA(P+Q)+d=0,
$$
i.e., there exists $\lambda\in\mathbb R$ such that
$$
-\lambda A(P+Q)=2\mathbf b.
$$
Let $R$ be the intersection of the conic with the line passing through $M$ and $V$ (see the graph in the accepted answer).
There exists $\mu$ such that $R=\mu M=\frac12\mu(P+Q)$. Then
$$
\frac14\mu^2(P+Q)^TA(P+Q) + \mu\mathbf b^T(P+Q)+c=0.
$$
Using $(1)$ we get
$$
\frac14\mu^2\big(P+Q\big)^TA\big(P+Q\big)
   + \big(\mu-\frac12\big)\mathbf b^T\big(P+Q\big)=0.
$$
Note that $\mu=\frac12$ implies that the conic is a parabola because, in this case,
$$
(P+Q)^TA(P+Q)=0
$$
and so $\det A=0$.
When the conic is not a parabola define
$$
\lambda = \frac{\mu^2}{2\mu-1}.
$$
Then
\begin{equation*}
(P+Q)^T\big(\lambda A(P+Q)+2\mathbf b\big)=0.
\end{equation*}
To complete the proof we need to find another vector, linearly independent from $P+Q$, and perpendicular to $\lambda A(P+Q)+2\mathbf b$. But
\begin{align*}
(P-Q)^T(\lambda A(P+Q) + 2\mathbf b) &= \lambda(P-Q)^TA(P+Q)
     + 2(P-Q)^T\mathbf b\\
    &= \lambda P^TAP-\lambda Q^TAQ + 2(P^T\mathbf b - Q^T\mathbf b)\\
    &= \lambda c - \lambda c + 2(-c + c)    &&\textrm{; by }(1)\\
    &= 0.
\end{align*}
