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In the definition of an ellipse, Definition ellipsedefn, we fixed two points called foci and looked at points whose distances to the foci
always added to a constant distance \(d\). Those prone to syntactical tinkering may wonder what, if any, curve we’d generate if we
replaced added with subtracted. The answer is a hyperbola.
Given two distinct points \(F_1\) and \(F_2\) in the plane and a fixed distance \(d\), a hyperbola is the set of all points \((x, y)\) in the plane such that
the absolute value of the difference of each of the distances from \(F_1\) and \(F_2\) to \((x, y)\) is \(d\). The points \(F_1\) and \(F_2\) are called the foci of the
hyperbola.
In the GeoGebra interactive below, adjusting the sliders for the points \(A\) and \(B\) trace points along the hyperbola
where
\[ \left | \text {the distance from $F_{1}$ to $A$} - \text {the distance from $F_{2}$ to $A$} \right | = 1 \]
and
\[ \left | \text {the distance from $F_{1}$ to $B$} - \text {the distance from $F_{2}$ to $B$} \right | = 1 \]
Note that the hyperbola has two parts, called branches. The center of the hyperbola is the midpoint of the line segment
connecting the two foci. The transverse axis of the hyperbola is the line segment connecting two opposite ends of the
hyperbola which also contains the center and foci. The vertices of a hyperbola are the points of the hyperbola which lie on the
transverse axis.
In addition, we will show momentarily that the hyperbola has a pair of asymptotes which the branches of the hyperbola
approach for large \(x\) and \(y\) values. They serve as guides to the graph. Schematically:
Figure 1: A hyperbola with center \(C\); foci \(F_1\), \(F_2\); and vertices \(V_1\), \(V_2\) and asymptotes (dashed)
Before we derive the standard equation of the hyperbola, we need to discuss one further parameter, the conjugate axis of
the hyperbola. The conjugate axis of a hyperbola is the line segment through the center which is perpendicular to the
transverse axis and has the same length as the line segment through a vertex which connects the asymptotes.
Schematically:
Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse
and conjugate axes whose center is the center of the hyperbola and whose diagonals are contained in the asymptotes.
This guide rectangle, much akin to the one we saw Section Ellipses to help us graph ellipses, will aid us in graphing
hyperbolas.
Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall assume that the center is \((0,0)\), the
vertices are \((a,0)\) and \((-a,0)\) and the foci are \((c,0)\) and \((-c,0)\). We’ll label the endpoints of the conjugate axis \((0,b)\) and \((0,-b)\). (Although \(b\) does
not enter into our derivation, we will justify this choice later.) As before, we assume \(a\), \(b\), and \(c\) are all positive
numbers.
The GeoGebra interactive below not only provides us with a detailed diagram of our generic hyperbola, but also a slider for
the distance parameter, \(d\). Adjusting \(d\) shows us how the shape of the hyperbola changes (as determined by the values of \(a\) and \(b\))
with fixed foci at \((c,0)\) and \((-c,0)\).
Since \((a,0)\) is on the hyperbola, it must satisfy the conditions of Definition hyperboladefn. That is, the distance from \((-c,0)\) to \((a,0)\) minus the distance from \((c,0)\)
to \((a,0)\) must equal the fixed distance \(d\). Since all these points lie on the \(x\)-axis, we get
\[ \begin{array}{rclr} \mbox {distance from $(-c,0)$ to $(a,0)$} - \mbox {distance from $(c,0)$ to $(a,0)$} & = & d & \\ (a+c) - (c-a) & = & d & \\ 2a & = & d \\ \end{array}\]
In other words, the fixed distance \(d\) from the definition of the hyperbola is actually the length of the transverse axis! (Where
have we seen that type of coincidence before?) Now consider a point \((x,y)\) on the hyperbola. Applying Definition hyperboladefn, we
get
Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation standardellipse,
we arrive at the following. (It is a good exercise to actually work this out.)
What remains is to determine the relationship between \(a\), \(b\) and \(c\). To that end, we note that since \(a\) and \(c\) are both positive numbers
with \(a < c\), we get \(a^2 < c^2\) so that \(a^2 - c^2\) is a negative number. Hence, \(c^2 - a^2\) is a positive number. For reasons which will become clear soon, we solve
the equation for \(\frac {y^2}{x^2}\):
As \(|x| \rightarrow \infty \), (Recall this means we are analyzing the behavior as \(x \rightarrow \infty \) and as \(x \rightarrow -\infty \).) the quantity \(\frac {\left (c^2 - a^2\right )}{x^2} \rightarrow 0\) so that \(\frac {y^2}{x^2} \approx \frac {\left (c^2 - a^2\right )}{a^2}\). By setting
\(b^{2} = c^{2} - a^{2}\) we get \(\frac {y^2}{x^2} \approx \frac {b^2}{a^2}\). This shows that \(y \approx \pm \frac {b}{a} x\), so that \(y = \pm \frac {b}{a} x\) are the asymptotes to the graph as predicted and our choice of labels
for the endpoints of the conjugate axis is justified. In our equation of the hyperbola we can substitute \(a^2 - c^2 = -b^2\) which
yields
The equation above is for a hyperbola whose center is the origin and which opens to the left and right. If the hyperbola were
centered at a point \((h,k)\), we would get the following.
The Standard Equation of a Horizontal (That is, a hyperbola whose branches open to the left and right)
Hyperbola
For positive numbers \(a\) and \(b\), the equation of a horizontal hyperbola with center \((h,k)\) is
If the roles of \(x\) and \(y\) were interchanged, then the hyperbola’s branches would open upwards and downwards and we would get
a ‘vertical’ hyperbola.
The Standard Equation of a Vertical Hyperbola
For positive numbers \(a\) and \(b\), the equation of a vertical hyperbola with center \((h,k)\) is:
The values of \(a\) and \(b\) determine how far in the \(x\) and \(y\) directions, respectively, one counts from the center to determine the guide
rectangle. In both cases, the distance from the center to the foci, \(c\), as seen in the derivation, can be found by the formula \(c = \sqrt {a^2 + b^2}\).
Lastly, note that we can quickly distinguish the equation of a hyperbola from that of a circle or ellipse because the
hyperbola formula involves a difference of squares where the circle and ellipse formulas both involve the sum of
squares.
\(~\)
Graph each of the following equations below in the \(xy\)-plane. Find the center, the lines which contain the
transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.
\(25(x-2)^2 - 4y^2 = 100\).
\(9y^2-x^2-6x=10\).
Graph \(f(x) = \sqrt {x^2-2x-3}\).
Find the standard form of the equation of a hyperbola which satisfies the following characteristics:
the asymptotes are \(y= \pm 2x\) and the vertices are \((\pm 5,0)\).
the hyperbola graphed below:
Owing to the difference of squares in \(25(x-2)^2 - 4y^2 = 100\), we work towards putting this equation into the form of Equation standardhhyperbola
or Equation standardvhyperbola. To that end, we rewrite \(y^2\) as \((y-0)^2\) and divide through by \(100\):
We identify \(h = 2\) and \(k = 0\), so the hyperbola is centered at \((2,0)\). We also see \(a = 2\), and \(b=5\), which means we move \(2\) units to
the left and to the right of the center and \(5\) units up and down from the center to arrive at points on the
guide rectangle: \((2-2, 0) = (0,0)\), \((2+2, 0) = (4,0)\), \((2, 0+5) = (2,5)\), and \((2, 0-5) = (2,-5)\). Since slant asymptotes pass through the center of the hyperbola as well as the
corners of the rectangle, we get the set-up as drawn below.
Since the \(y^2\) term is being subtracted from the \(x^2\) term, we are in the situation of Equation standardhhyperbola. Hence, the branches of the
hyperbola open to the left and right so the transverse axis lies along the \(x\)-axis and the conjugate axis lies along the
vertical line \(x = 2\).
Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices
are \((0,0)\) and \((4,0)\). To find the foci, we need \(c = \sqrt {a^2 + b^2} = \sqrt {4+25} = \sqrt {29}\). Since the foci lie on the transverse axis, we move \(\sqrt {29}\) units to the left and right of \((2,0)\)
to arrive at \((2 - \sqrt {29},0)\) (approximately \((-3.39, 0)\)) and \((2 + \sqrt {29}, 0)\) (approximately \((7.39, 0)\)).
Lastly, to determine the equations of the asymptotes, recall that the asymptotes pass through the center of the
hyperbola, \((2,0)\), as well as the corners of guide rectangle. As such, they have slopes of \(\pm \frac {b}{a} = \pm \frac {5}{2}\). Feeding this information into
the point-slope equation of a line, Equation pointslope, we get \(y -0 = \pm \frac {5}{2} (x - 2)\), so the asymptotes are \(y = \frac {5}{2}x - 5\) and \(y = -\frac {5}{2}x + 5\). Putting it all together, we get
our final graph below.
Since we have a difference of squares in \(9y^2-x^2-6x=10\), we aim to transform our given equation into Equation standardhhyperbola or Equation standardvhyperbola. As
we’ve seen with the other conic sections, we begin with completing the square.
Now that this equation is in the standard form of Equation standardvhyperbola, we identify \(h = -3\) and \(k = 0\) so the center is \((-3,0)\). We also see so \(a=1\),
and \(b=\frac {1}{3}\) which means that we move \(1\) unit to the left and to the right of the center and \(\frac {1}{3}\) units up and down from the
center to arrive at points on the guide rectangle: \((-3-1,0) = (-4,0)\), \((-3+1,0) = (-2,0)\), \(\left (-3, 0+\frac {1}{3} \right ) =\left (-3, \frac {1}{3} \right )\) and \(\left (-3, 0-\frac {1}{3} \right ) =\left (-3, -\frac {1}{3} \right )\).
Since the \(x^2\) term is being subtracted from the \(y^2\) term, we know the branches of the hyperbola open upwards and
downwards. This means the transverse axis lies along the vertical line \(x=-3\) and the conjugate axis lies along the \(x\)-axis.
As a result, we get the vertices are \(\left (-3, \frac {1}{3}\right )\) and \(\left (-3, -\frac {1}{3}\right )\).
To find the foci, we use \(c = \sqrt {a^2 + b^2} = \sqrt {\frac {1}{9} + 1} = \frac {\sqrt {10}}{3}\). Since the foci lie on the transverse axis, we move \(\frac {\sqrt {10}}{3}\) units above and below \((-3, 0)\) to arrive at \(\left (-3, \frac {\sqrt {10}}{3}\right )\)
and \(\left (-3, -\frac {\sqrt {10}}{3}\right )\).
To determine the asymptotes, we use the fact the asymptotes pass through the center of the hyperbola, \((-3,0)\), as well
as the corners of guide rectangle, so they have slopes of \(\pm \frac {b}{a} = \pm \frac {1}{3}\). Once again we use the point-slope equation of a line,
Equation pointslope, to get the two asymptotes \(y = \frac {1}{3}x + 1\) and \(y = -\frac {1}{3}x - 1\). Our final graph is below.
Graphing \(f(x) = \sqrt {x^2-2x-3}\) amounts to graphing the equation \(y= \sqrt {x^2-2x-3}\). In order to use the tools we’ve learned in this chapter, we first square
both sides to get a quadratic equation in two variables: \(y^2 = (\sqrt {x^2-2x-3})^2\). We get \(y^2 = x^2-2x-3\) or \(y^2 - x^2 + 2x = -3\). We now set about transforming this equation into
the form stated in Equation standardhhyperbola or Equation standardvhyperbola.
We get the equation into the form of Equation standardhhyperbola and identify \(h=1\) and \(k=0\) so the center is \((1,0)\). We have \(a = b = 2\), which means we move \(2\)
units to the left, to the right, up and down from the center to find points on the guide rectangle: \((1-2,0) = (-1,0)\), \((1+2, 0) = (3,0)\), \((1,0-2) = (1,-2)\) and \((1,0+2) = (1,2)\). Of these four
points, the vertices are \((-1,0)\) and \((3,0)\) since the hyperbola opens to the left and to the right. As usual, we the
guide rectangle helps us sketch the hyperbola along with its slant asymptotes, which we find are \(y - 0 = \pm (x-1)\) or \(y = x-1\) and
\(y = -x+1\).
We know since \(f\) is a function, the graph of \(f\) cannot be the entire hyperbola, otherwise the graph would fail the vertical
line test. Since, by definition, \(\sqrt {x^2-2x-3} \geq 0\), we know \(f(x) \geq 0\). Hence the graph of \(f\) is the upper half of the hyperbola, as shown
below.
Plotting the data given to us below, we know the branches of the hyperbola open to the left and to the
right. This means the our answer will take the form of Equation standardhhyperbola.
Since the center is the midpoint of the vertices, we see the center is \((0,0)\), so \(h = k = 0\). Moreover, since the
vertices are exactly \(5\) units from the center, we know \(a=5\) so \(a^2 = 25\). All that remains to find is the value of
\(b^2\).
Recall that the slopes of the asymptotes are \(\pm \frac {b}{a}\). Since \(a = 5\) and the slope of the line \(y=2x\) is \(2\), we have that \(\frac {b}{5} = 2\), so \(b=10\). Hence, \(b^2 = 100\). Our
final answer is \(\frac {x^2}{25} - \frac {y^2}{100} = 1\).
From what we are given on the graph, the equation of the hyperbola takes the form of Equation standardvhyperbola. The vertices
appear to be \((3,1)\) and \((3,-3)\) whose midpoint gives us the center as \((3,-1)\). Hence, \(h=3\) and \(k =-1\). Moreover, since the vertices
are \(2\) units above and below the center, we know \(b = 2\) so \(b^2 = 4\). All that remains is for us to find the value of
\(a^2\).
Since we are given two additional points, \((0,3)\) and \((0,-5)\), we choose one of them, \((0,3)\) to find \(a^2\) and use the other, \((0,-5)\) to partially
check our answer.
At this stage, we know the equation of the hyperbola is
To Write the Equation of a Hyperbola in Standard Form
Group common variables together on one side of the equation and put the constant on the other.
Complete the square on both variables as needed.
Divide both sides, if needed, to obtain \(1\) on one side of the equation.
Hyperbolas can be used in so-called ‘ trilateration,’ or ‘positioning’ problems. The procedure outlined in the next example is
the basis of the (now defunct) LOng Range Aid to Navigation ( LORAN for short) system. (GPS now rules the positioning
kingdom. Is there still a place for LORAN? Do satellites ever malfunction?)
\(~\)
Jeff is stationed \(10\) miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive
Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call \(9\) seconds earlier than Carl. If the speed of
sound that night is \(760\) miles per hour, determine a hyperbolic path along which Sasquatch must be located.
By a stroke of luck, Kai is also camping in the woods at this time. He is \(6\) miles due north of Jeff and heard the
Sasquatch call \(18\) seconds after Jeff did. Use this added information to locate Sasquatch.
Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is \(760\) miles per
hour, we can determine how much closer Sasquatch is to Jeff by multiplying
This means that Sasquatch is \(1.9\)
miles closer to Jeff than it is to Carl. In other words, Sasquatch must lie on a path where
\[\text {(the distance to Carl)} - \text {(the distance to Jeff)} = 1.9\]
This is exactly the
situation in the definition of a hyperbola, Definition hyperboladefn. In this case, Jeff and Carl are located at the foci, (We
usually like to be the center of attention, but being the focus of attention works equally well.) and our fixed
distance \(d\) is 1.9. For simplicity, we assume the hyperbola is centered at \((0,0)\) with its foci at \((-5, 0)\) and \((5, 0)\).
The GeoGebra interactive below displays all of the available data. Adjusting the slider allows us to trace the
possible location of the Sasquatch along the western (left) branch of the hyperbola.
We are seeking a curve of the form \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) in which the distance from the center to each focus is \(c = 5\). As we saw in the derivation
of the standard equation of the hyperbola, Equation standardhhyperbola, \(d = 2a\), so that \(2a = 1.9\), or \(a = 0.95\) and \(a^2 = 0.9025\). All that remains is to find \(b^2\). To that end, we recall
that \(a^2 + b^2 = c^2\) so \(b^2 = c^2 - a^2 = 25 - 0.9025 = 24.0975\). Since Sasquatch is closer to Jeff than it is to Carl, it must be on the western (left hand) branch of
Since Jeff is positioned at \((-5, 0)\), we place Kai at \((-5, 6)\). This puts the center of the new hyperbola at \((-5, 3)\). Plotting Kai’s position and
the new center gives us the diagram below on the left.
The second hyperbola is vertical, so it must be of the form \(\frac {(y-3)^2}{b^2} - \frac {(x+5)^2}{a^2} = 1\). As before, the distance \(d\) is the length of the major axis,
which in this case is \(2b\). We get \(2b = 3.8\) so that \(b = 1.9\) and \(b^2 = 3.61\). With Kai \(6\) miles due North of Jeff, we have that the distance from the center
to the focus is \(c = 3\). Since \(a^2 + b^2 = c^2\), we get \(a^2 = c^2 - b^2 = 9 - 3.61 = 5.39\).
Kai heard the Sasquatch call after Jeff, so Kai is farther from Sasquatch than Jeff. Thus Sasquatch must lie on the
southern branch of the hyperbola
Using GeoGebra, we find exactly one point which lies both on the western branch of the hyperbola determined by Jeff
and Carl along with the southern branch of the hyperbola determined by Kai and Jeff. We find (To determine the
coordinates of this point of intersection exactly, we would need techniques for solving systems of non-linear equations
(which we won’t see until Section NonLinearEquations), so we use a graphing utility, in this case, GeoGebra.) Sasquatch was at
approximately at \((-0.9629, -0.8113)\) when it called.
Note that at this point, the Sasquatch is approximately \(6.02\) miles from Carl, \(4.12\) miles from Jeff, and \(7.92\) miles from Kai. We check
that \(|6.02 - 4.12| = 1.9\) and \(|4.12 - 7.92| = 3.8\), proving the point \((-0.9629, -0.8113)\) is indeed on both hyperbolas.
Each of the conic sections we have studied in this chapter result from graphing equations of the form \(Ax^2 + Cy^2 + Dx + Ey + F = 0\) for different choices of \(A\),
\(C\), \(D\), \(E\), and (See Section PolarConics to see why we skip \(B\).)\(F\). While we’ve seen examples demonstrate how to convert an equation from
this general form to one of the standard forms, we close this chapter with some advice about which standard form to choose.
(We formalize this in Exercise conicsclassificationnoxytermex.)
Strategies for Identifying Conic Sections
Suppose the graph of equation \(Ax^2 + Cy^2 + Dx + Ey + F = 0\) is a non-degenerate conic section. (That is, a parabola, circle, ellipse, or hyperbola – see
Section IntrotoConics.)
If just one variable is squared, the graph is a parabola. Rewrite the equation in the standard form given in
Equation standardvparabola (if \(x\) is squared) or Equation standardhparabola (if \(y\) is squared).
If both variables are squared, look at the coefficients of \(x^2\) and \(y^2\), \(A\) and \(C\).
If \(A=C\), the graph is a circle. Rewrite the equation in the standard form given in Equation standardcircle.
If \(A \neq C\) but \(A\) and \(C\) have the same sign, the graph is an ellipse. Rewrite the equation in the standard form given in
Equation standardellipse.
If \(A\) and \(C\) have the different signs, the graph is a hyperbola. Rewrite the equation in the standard form given in
either Equation standardhhyperbola or Equation standardvhyperbola.
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