hyperbolaHyperbola22007-05-21 12:16:142009-04-22 15:07:11Topicfamous curveshyperbolafocusfocifocal radiusfocal radiiapex of hyperbolaapices of hyperbolatransversal axisconjugate axisrectangular hyperbola% this is the default PlanetMath preamble. as your knowledge
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\PMlinkescapeword{contains} \PMlinkescapeword{branches}
A {\em hyperbola} is the locus of points $P$ in the Euclidean plane such that the distances of $P$ from two \PMlinkescapetext{fixed points} (the foci $F_1$ and $F_2$) differ from each other by a constant amount ($\pm2a$).\, The line segments connecting a point of the hyperbola to the foci are called {\em focal radii}.
We obtain the simplest equation for the hyperbola by choosing the foci on the other coordinate axis and equidistant ($= c > a > 0$) from the origin.\; Let\; $F_1 = (-c,\,0)$\, and $F_2 = (c,\,0)$.\, Then the locus condition for the point\, $P = (x,\,y)$\, of the hyperbola is
$$\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2} = \pm2a.$$
The simplifying of this, via two squarings, yields the equation of the hyperbola
\begin{align}
\frac{x^2}{a^2}\!-\!\frac{y^2}{b^2} \;=\; 1.
\end{align}
Here we have denoted\, $c^2\!-\!a^2 = b^2$\, where\, $b > 0$.
Since the equation (1) contains only the squares of $x$ and $y$ we can infer that the hyperbola is symmetric with respect to the coordinate axes and the origin; this is naturally clear on grounds of the definition of hyperbola, too.
Solving (1) for $y$ we get
\begin{align}
y \;=\; \pm\frac{b}{a}\sqrt{x^2\!-\!a^2}.
\end{align}
This shows that $y$ is real only for\, $x \geqq a$;\, for\, $x = \pm a$\, we have\, $y = 0.$\, When
$|x|$ increases from $a$ to infinity, $|y|$ increases from $0$ to infinity.\, So we see that the hyperbola consists of two distinct branches from which the one is to the \PMlinkescapetext{right} from the line\, $x = a$\, and the other to the left from the line\, $x = -a$.\, These lines touch the branches at the points\, $(a,\,0)$\, and\, $(-a,\,0)$,\, which are called the {\em apices} of the hyperbola.
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The line segment connecting the apices is the {\em transversal axis} of the hyperbola.\, The line segment on the $y$-axis from $-b$ to $b$ is the {\em conjugate axis} of the hyperbola.\, By the Pythagorean theorem, the equation\,
$b^2 = c^2-a^2$\, shows that the distance between an end of the transversal and an end of the conjucate axis is equal to $c$.
Let's consider the part
$$y = \frac{b}{a}\sqrt{x^2-a^2}$$
of the hyperbola situated in the first quadrant ($x > a$) and the line
$$y \;=\; \frac{b}{a}x.$$
The difference of their ordinates corresponding a same abscissa $x$ may be written
$$\Delta \;:=\; \frac{b}{a}(x\!-\!\sqrt{x^2\!-\!a^2}) \;=\; \frac{ab}{x\!+\!\sqrt{x^2\!-\!a^2}}\;\;\;(> 0).$$
But\, $\Delta \to 0$\, as\, $x \to \infty$,\, whence this \PMlinkescapetext{branch} of the hyperbola approaches unlimitedly from below the line.\, Accordingly, the line\, $y = \frac{b}{a}x$\, is an asymptote of our curve.\, By the symmetry, the hyperbola (1) has two asymptotes
\begin{align}
y \;=\; \pm\frac{b}{a}x.
\end{align}
The asymptotes are easy to draw, since they are the lengthened diagonals of the rectangle whose sides are on the lines\, $x = \pm a$\, and\, $y = \pm b$.\, The hyperbola may be sketched by utilising that rectangle and the asymptotes.
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Both asymptotes form with the transversal axis an angle whose tangent is equal to $\frac{b}{a}$.\, This equals 1, when the transversal axis and the conjugate axis are equal ($a = b$); then the rectangle is a square and one speaks of a {\em rectangular hyperbola}. See also the entry transition to skew-angled coordinates.
The tangent line of the hyperbola (1) is
$$\frac{x_0x}{a^2}-\frac{y_0y}{b^2} = 1,$$
where\, $(x_0,\,y_0)$\, is the point of tangency on the hyperbola (see tangent of conic section).\, The tangent halves the angle between the focal radii drawn from\, $(x_0,\,y_0)$.
\begin{thebibliography}{8}
\bibitem{LL}{\sc L. Lindel\"of}: {\em Analyyttisen geometrian oppikirja}.\, Kolmas painos.\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924).
\end{thebibliography}