Hyperbolas in Algebra
These are my notes on hyperbolas in algebra.
Equations and Graphs
The third type of conic section is a hyperbola. A hyperbola is the set of points
in a plane, the differences of whose distances from two fixed points is
constant. Each fixed point is called a focus of the hyperbola.
When looking at a hyperbola a point is shown on a hyperbola with distance from
one focus and the distance from the second focus. Regardless of the location of
the point, d2-d1=2a. The transverse axis is the line segment connecting the
vertices and its length equals 2a.
The two parts of the hyperbola are the left branch and the right branch. It can
also be labeled the upper and lower branch. A line segment connecting the points
is the conjugate axis. The lines are asymptotes for each respective hyperbola.
They can be used as an aid in graphing. The dashed rectangle is sometimes called
the fundamental rectangle.
By the vertical line test, a hyperbola cannot be represented by a function, but
many can be described by the following equations. The constants a,b, and c are
positive.
The hyperbola with center at the origin, horizontal transverse axis, and
equation:
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
has asymptotes \(y = \pm \frac{b}{a} x\), vertices \((\pm a,0)\), where
\(c^2=a^2+b^2\). The hyperbola with center at its origin, vertical transverse
axis, and equation:
\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]
has asymptotes \(y = \pm \frac{a}{b} x\), vertices \((0, \pm a)\), and foci
\((0, \pm c)\), where \(c^2 = a^2+b^2\).
A hyperbola consists of two solid curves or branches. The asymptotes, foci,
transverse axis, and fundamental rectangle are not part of the hyperbola, but
are aids for sketching its graph.
Example 1
Sketching the graph of a hyperbola
Sketch a graph of \(\frac{x^2}{4} - \frac{y^2}{9} = 1\)
Solution:
The equation is in standard form with a=2 and b=3. It has a horizontal
transverse axis with vertices \((\pm 2,0)\). The endpoints of the conjugate axis
are \((0, \pm 3)\). To locate the foci find c.
\[c^2=a^2+b^2=4+9=13\]
\[c=\sqrt{13} = 3.61\]
The foci are \((\pm \sqrt{13}, 0)\). The asymptotes are \(y=\pm \frac{b}{a} x\)
or \(y=\pm \frac{3}{2} x\).
Example 2
Finding the equation of a hyperbola
Find the equation of the hyperbola centered at the origin with a vertical
transverse axis of length 6 and focus (0,5).
Solution:
Since the hyperbola is centered at the origin with a vertical axis, its equation
is \(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\). The transverse axis has length 6=2a,
so a=3. Since one focus is located at (0,5), c=5. We can find b by using the
following equation:
\[b^2=c^2-a^2\]
\[b=\sqrt{c^2-a^2}\]
\[b=\sqrt{5^2-3^2}=4\]
The equation of this hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}=1\). Its
asymptotes are \(y=\pm \frac{a}{b} x\) or \(y=\pm \frac{3}{4} x\).
Example 3
Finding the equation of a hyperbola
Find the standard equation of the hyperbola. Identify the vertices, foci, and
asymptotes.
Solution:
Because the hyperbola is centered at (0,0) with a horizontal transverse axis,
its equation has the form:
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]
The asymptotes are given by \(y=\pm \frac{b}{a}\). The fundamental rectangle is
determined by the four points \((\pm 4,0)\) and \((0,\pm 2)\), and its diagonals
correspond to the asymptotes. It follows that a=4 and b=2. Thus the standard
equation of the hyperbola is:
\[\frac{x^2}{16}-\frac{y^2}{4}=1\]
The vertices are \((\pm 4,0)\), and the asymptotes are \(y=\pm \frac{1}{2} x\).
To find the coordinates of the foci, find c:
\[c^2=a^2+b^2=4^2+2^2=20\]
\[c=\sqrt{20}=4.47\]
The coordinates of the foci are \((\pm \sqrt{20},0)\)
Reflective Properties
Hyperbolas have an important reflective property. If a hyperbola is rotated
about the x-axis, a hyperboloid is formed. Any beam of light that is directed
toward focus, will be reflected by the hyperboloid toward the other focus.
Telescopes sometimes make use of both parabolic and hyperbolic mirrors. When
parallel rays of light from distant stars strike the large parabolic mirror,
they are reflected toward its focus. A smaller hyperbolic mirror is placed so
that its focus is also located at the first focus. Light rays striking the
hyperbolic mirror are reflected toward is other focus, through a small hole in
the parabolic mirror.
Translations of hyperbolas
If the equation of a hyperbola is either \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) or
\(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\), then the center of the hyperbola is
(0,0). We can use the translations of graphs to find the equation of a hyperbola
centered at (h,k) by replacing x with (x-h) and y with (y-k). The constants a,b,
and c are positive.
Example 4
Graphing a hyperbola with center (h,k)
Graph the hyperbola whose equation \(\frac{(y+2)^2}{9}-\frac{(x-2)^2}{16}=1\).
Solution:
The hyperbola has a vertical transverse axis and its center is (2,-1). Since
\(a^=9\) and \(b^2=16\), it follows that \(c^2=a^2+b^2=9+16=25\). Thus a=3, b=4,
and c=5. The vertices are located 3 units above and below the center of the
hyperbola and the foci are located 5 units above and below the center of the
hyperbola. The vertices are \((2,-1 \pm 3)\) or (2,1) and (2,-5), and the foci
are \((2,-1 \pm 5)\) or (2,3) and (2,-7). The asymptotes are given by:
\[y=\pm \frac{a}{b} (x-h)+k\]
\[y=\pm \frac{3}{4} (x-2)-2\]
Example 5
Finding the standard equation of the hyperbola
Write \(9x^2-18x-4y^2-16y=43\) in the standard form for a hyperbola centered at
(h,k).
Solution:
We can write the given equation in standard form by completing the square.
\[9x^2-18x-4y^2-16y=43\]
\[9(x^2-2x+1)-4(y^2+4y+4)=43+9-16\]
\[9(x-1)^2-4(y+2)^2=36\]
\[\frac{(x-1)^2}{4}-\frac{(y+2)^2}{9}=1\]
The center is (1,-2). Because a=2 and the transverse axis is horizontal, the
vertices of the hyperbola are \((h \pm a,k)=(1 \pm 2,-2)\).
Solving systems of nonlinear equations
If the equation of a hyperbola occurs in a system of equations, then the system
of equations is nonlinear.
Example 6
Solving a system of nonlinear equations
\[4x^2-9y^2=36\]
\[9x^2+25y^2=262\]
Solution:
To eliminate the x-variable, multiply the first equation by -9, the second
equation by 4, and then add the resulting equations.
\[-36x^2+81y^2=-324\]
\[36x^2+100y^2=1048\]
Add these:
\[181y^2=724\]
Now solve:
\[y^2=\frac{724}{181}=4\]
\[y=\pm 2\]
To determine the corresponding x-values, let \(y=\pm 2\) in the given equation
\(4x^2-9y^2=36\).
\[4x^2-9(\pm 2)^2=36\]
\[4x^2-36=36\]
\[4x^2=72\]
\[x^2=18\]
\[x=\pm \sqrt{18}\]
\[x=\pm 3 \sqrt{2}\]
There are four solutions to this system of nonlinear equations:
\[(\pm 3\sqrt{2}, \pm 2)\]