Exercise 5.6.3 (Intersections of a line with an hyperbola)

Question

Apply the analysis in the text to the hyperbola $x^2 - y^2 = 1$ with $(x_0, y_0) = (1,0)$, and thus find the slope $m_0$ of the tangent line, and the slopes $m_1$ and $m_2$ that give no second intersection.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put $f(x,y) = x^2 - y^2 - 1$. The equation of the tangent line $T$ at the point $(x_0, y_0) = (1,0)$ is given by 
\begin{align*}
0 &= \frac{\partial f}{\partial x}(x_0,y_0) (x- x_0) +  \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0),
\end{align*}
where
\begin{align*}
\frac{\partial f}{\partial x}(x_0,y_0) = 2x_0 = 2 ,\qquad \frac{\partial f}{\partial y}(x_0,y_0) = -2 y_0 =  0.
\end{align*}
Therefore the equation of $T$ is $2(x-1) = 0$, or $x = 1$. The slope $m_0$ of $T$ is $m_0 = \infty$.

A non-vertical line $D$ passing trough $(1,0)$ with slope $m$ has equation
$$D: y = m (x-1).$$
Then the abscissas of the points of intersection of $D$ with the hyperbole are the roots of
\begin{align*}
   p(x) &= f(x, m(x-1))\
&= x^2 - m^2(x-1)^2 - 1\ 
&= (x-1)[ x+1 - m^2(x-1)]\
&= (x-1)[x(1-m^2) + m^2 + 1].
\end{align*}
Then
\begin{align*}
p(1+u) &= u [(1+u)(1-m^2) + m^2 + 1]\
&= 2u + u^2 (1-m^2),
\end{align*}
thus the multiplicity of intersection is always $M = 1$ (because the tangent line is vertical).

If $m
e \pm1$, then the line $D$ contains a second point of intersection, given by $x = (m^2 + 1)/(m^2 - 1)$ (and $y = m(x-1)$).

Therefore the slopes $m_1$ and $m_2$ that give no second intersection are $m_1 = 1, m_2 = -1$ (the two parallels to the asymptotes passing through $(1,0)$).
\end{proof}

Exercise 5.6.3 (Intersections of a line with an hyperbola)

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Exercise 5.6.3 (Intersections of a line with an hyperbola)

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