Hyperbolic local replacement in the neighborhood of a curve's vertex

Question

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

1. both lines are tangent to the hyperbola at $p$ and $q$ and
2. 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.

Answer 1

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 $B(o,\epsilon)$.

Answer 2

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*}

Answer 3

Construction

As shown in the other two answers, any hyperbola with equation $$ (\mathbf x^T+\lambda M^2)A(\mathbf x+\lambda M)+d=0 $$ passing through $P$ and $Q$ will have the desired properties. In consequence, the equation of the tangent line at a point $Z$ is $$ (Z^T+\lambda M^T)A\mathbf x + \lambda M^TA(Z + \lambda M) + d=0 $$ i.e., $$ (Z^T+\lambda M^T)A(\mathbf x + \lambda M) + d = 0.\tag1 $$ With $$ A = \begin{pmatrix} a &c\\ c &b \end{pmatrix} $$ we have four unknowns $a, b, c, d$ and two linear equations: \begin{align*} (P^T+\lambda M^T)A(P+\lambda M) + d &= 0\\ (Q^T+\lambda M^T)A(Q+\lambda M) + d &= 0, \end{align*} plus the equations $(1)$ for the tangents at $P$ and $Q$ passing through $V=(0,0)$ \begin{align} \lambda(P^T+\lambda M^T)AM + d &= 0\\ \lambda(Q^T+\lambda M^T)AM + d &= 0. \end{align} Then the first two become \begin{align} (P^T+\lambda M^T)AP &= 0\tag2\\ (Q^T+\lambda M^T)AQ &= 0 \end{align} and their subtraction gives \begin{equation} Q^TAQ = P^TAP\tag3. \end{equation} Thus, we are left with two equations, namely $(2)$ and $(3)$, represented, in matrix form, as $$ \begin{pmatrix} (x_1+\lambda s)x_1 &(y_1+\lambda t)y_1 &(x_1+\lambda s)y_1 + (y_1+\lambda t)x_1\\ x_2^2-x_1^2 &y_2^2-y_1^2 &2x_2y_2-2x_1y_1 \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $$ where $(x_1,y_1)=P$, $(x_2,y_2)=Q$ and $(s,t)=M=\frac12(x_1+x_2,y_1+y_2)$. Replacing $f_1\leftarrow f_1+f_2$: $$ \begin{pmatrix} x_2^2+\lambda sx_1 &y_2^2+\lambda ty_1 &2x_2y_2+\lambda sy_1 + \lambda tx_1\\ x_2^2-x_1^2 &y_2^2-y_1^2 &2x_2y_2-2x_1y_1 \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}.\tag4 $$ If $x_1+x_2\ne0$, we can apply a rotation of angle $$ \theta=-\operatorname{arccot}\frac{y_1+y_2}{x_1+x_2} $$ and reduce ourselves to the case where $x_2=-x_1$. Let momentarily denote by $\bar x_1$ etc. the rotated variables. Interchanging $\bar x_1$ with $\bar x_2$ if necessary, choose $\bar x_1$ to have the sign of $$ \det\begin{pmatrix} x_1 &x_2\\ y_1 &y_2 \end{pmatrix}. $$ Thus, the rotation of angle $\theta$ will leave $\bar y_1$ and $\bar y_2$ in the same half-plane, i.e., we will have \begin{equation}\label{samesign} \operatorname{sg}(\bar y_1)=\operatorname{sg}(\bar y_2), \end{equation} as it can be easily verified from the equation $$ \begin{pmatrix} \phantom-\cos\theta &\sin\theta\\ -\sin\theta &\cos\theta \end{pmatrix} \begin{pmatrix} x_1 &x_2\\ y_1 &y_2 \end{pmatrix} = \begin{pmatrix} \bar x_1 &-\bar x_1\\ \bar y_1 &\phantom-\bar y_2 \end{pmatrix}. $$ In conclusion, we can assume $s=0$, $x_1\ne0$, $y_1+y_2\ne0$, with $y_1$ and $y_2$ having the same sign. Under these circumstances $(4)$ becomes $$ \begin{pmatrix} x_1^2 &y_2^2+\lambda ty_1 &\lambda tx_1-2x_1y_2\\ 0 &y_2^2-y_1^2 &-2x_1(y_1+y_2) \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $$ or $$ \begin{pmatrix} x_1^2 &y_2^2+\lambda ty_1 &-(2y_2-\lambda t)x_1\\ 0 &y_2-y_1 &-2x_1 \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $$ There are two cases:

{$y_1=y_2$} In this case $x_1,y_1\ne0$ and so the solution is generated by $$ (a,b,c)=((1+\lambda)y_1^2,-x_1^2,0), $$ with $\det A=-(1+\lambda)x_1^2y_1^2<0$.

{$y_1\ne y_2$} In this case the equations translate into $$ b=\frac{2x_1}{y_2-y_1}c\quad\textrm{and}\quad a = \Big(\frac{2y_2-\lambda t}{x_1} -2\frac{y_2^2+\lambda ty_1}{x_1(y_2-y_1)}\Big)c. $$ Then \begin{align*} \det A &= ab-c^2\\ &= \Big(2\frac{2y_2-\lambda t}{y_2-y_1} - 4\frac{y_2^2+\lambda ty_1}{(y_2-y_1)^2}\Big)c^2-c^2\\ &= \frac{2c^2}{(y_2-y_1)^2} \big((y_2-\lambda t)(y_2-y_1)-2(y_2^2+\lambda ty_1)\big)-c^2\\ &= -\frac{2c^2}{(y_2-y_1)^2}(y_1+y_2)(y_2+\lambda t)-c^2 \end{align*} and it is enough to show that $(y_1+y_2)(y_2+\lambda t) > 0$ or $$ t(y_2+\lambda t)>0, $$ which does hold because $y_1$ and $y_2$, and hence $t$, are of the same sign.

In both cases, the value of $d$ can be easily obtained.