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6.4: Hyperbolas

Created by: CK-12

Learning Objectives

  • Understand what results when a cone is intersected by a plane that intersects both sides of the cone.
  • Understand the focal-directrix property for hyperbolas.
  • Recognize and work with equations for hyperbolas.
  • Understand how a hyperbola can be described in terms of its asymptotes and foci.
  • Understand the equivalence of different definitions of hyperbolas.

Two-part Conic Sections

The final way of cutting a cone with a plane appears at first to be much messier. Not only does it result in an infinite shape, but there are two pieces that aren’t even connected! When the plane slices through two parts of the cone, the two infinite “U”-shaped parts are together called a hyperbola.

In this section we will see that this sprawling shape actually has some beautiful properties that make it as noble as its cousins.

The Focal Property

Even though this shape seems much harder to conceive of than an ellipse, the hyperbola has a defining focal property that is as simple as the ellipse’s. Remember, an ellipse has two foci and the shape can be defined as the set of points in a plane whose distances to these two foci have a fixed sum.

Hyperbolas also have two foci, and they can be defined as the set of points in a plane whose distances to these two points have the same difference. So in the picture below, for every point P on the hyperbola, |d_2 - d_1| = C for some constant C.

Review Questions

  1. The hyperbola is infinite in size. In mathematics this is called unbounded, which means no circle, no matter how large, can enclose the shape. Explain why a focal property involving a difference results in an unbounded shape, while a focal property involving a sum results in a bounded shape.

Review Answers

  1. Answer should include the following concept: In the case of an ellipse, we had two distances summing to a constant. Since the distances are both positive then there is a limit to the size of the numbers. In the case of hyperbolas, two very large positive numbers can have a much smaller difference.

To prove the focal property of hyperbolas, we examine Dandelin’s sphere construction. Unlike the construction for ellipses, which used two spheres on one side of the cone, and the sphere construction for parabolas, which used one sphere on one side of the cone, this construction uses two spheres, one on each side of the cone. As with the ellipse construction, each sphere touches the plane at one of the foci of the hyperbola. And as with the argument for the elliptical focal property, the argument uses the fact that tangents from a common point to a sphere are equal.

In the above diagram, suppose P is an arbitrary point on the hyperbola. We would like to examine the difference PF_2 - PF_1. Let C_1 be the point on the upper sphere that lies on the line between P and the vertex of the cone. Let C_2 be the point on the upper sphere when this line is extended (so P, C_1, and C_2 are all on the same line and PC_1 + C_1 C_2 = PC_2 and the cone By the common tangent property, PF_1 = PC_1 and PF_2 = PC_2 for some points C_1 and C_2 on the circles where the spheres meet the cone. So PF_2 - PF_1 = PC_2 - PC_1 = (PC_1 + C_1 C_2) - PC_1 = C_1 C_2. But C_1 C_2 is the distance along the cone between the two circles of tangency and is constant regardless of the choice of C_1 and C_2. So the difference PF_2 - PF_1 is constant.

Equations for Hyperbolas

To derive the equation for a hyperbola, we can’t do exactly what we did for ellipses. Remember, when we first derived the equations for ellipses, we were defining them as “stretched out circles.” Hyperbolas aren’t as obviously a simple dilation of a shape as basic as a circle. Instead, we’ll use the focal property to derive their equation. This was what we said could be done for ellipses when I said “simply read the proof of the focal property backwards.” To derive the equation form of hyperbolas from the focal property, I’ll actually show you the steps in the correct order.

Suppose we have a curve (actually a pair of curves), satisfying the focal property for hyperbolas. Let’s orient the hyperbola so that the two foci are on the y-axis, and equidistant from the origin. In the diagram below the foci are labeled with the points (0,c), and (0,-c).

The focal property states that the difference in distances between an arbitrary point P=(x,y) on the hyperbola and (0,c), and (0,-c) is a constant. In particular, we know that this constant can be computed from any point on the hyperbola, for instance the point on the y-axis marked (0,a). The distance between (0,a) and (0,-c) is a + c and the distance between (0,a) and (0,c) is c - a. The difference between these two quantities is (c + a) - (c - a) = 2a. So 2a, the distance between the two y-intercepts of the hyperbola, is the constant in the focal property for the hyperbola. Using this distance formula and the focal property, we have:

\sqrt{(x - 0)^2 + (y -(-c))^2} - \sqrt{(x - 0)^2 + (y - c)^2} &= 2a \\x^2 + (y + c)^2 - 2 \sqrt{x^2 + (y + c)^2} \sqrt{x^2 + (y - c)^2} + x^2 + (y - c)^2 &= 4a^2 \\2x^2 + (y + c)^2 + (y - c)^2 - 4a^2 &= 2 \sqrt{x^2 + (y + c)^2} \sqrt{x^2 + (y - c)^2} \\(2x^2 + 2y^2 + 2c^2 - 4a^2)^2 &= 4(x^2 + (y + c)^2)(x^2 + (y - c)^2) \\4x^4 + 8x^2 y^2 + 8c^2 x^2 + 8c^2 y^2 + 4y^4 - 16a^2 x^2 + 4c^4 - 16a^2 c^2 - 16a^2 y^2 + 16a^4 &= 4x^4 + 4x^2 (y + c)^2 + 4x^2 (y - c)^2 + 4(y + c)^2  (y - c)^2 \\4x^4 + 8x^2 y^2 + 8c^2 x^2 + 8c^2 y^2 + 4y^4 - 16a^2 x^2 + 4c^4 - 16a^2 c^2 - 16a^2 y^2 + 16a^4 &= 4x^4 + 4x^2 (y + c)^2 + 4x^2 (y - c)^2 + 4(y^2 - c^2)^2 \\4x^4 + 8x^2 y^2 + 8c^2 x^2 + 8c^2 y^2 + 4y^4 - 16a^2 x^2 + 4c^4 - 16a^2 c^2 - 16a^2 y^2 + 16a^4 &= 4x^4 + 8x^2 y^2 + 8c^2 x^2 + 4y^4 - 8c^2 y^2 + 4c^4 \\16c^2 y^2 - 16a^2 x^2 - 16a^2 c^2 - 16a^2 y^2 + 16a^4 &= 0 \\c^2 y^2 - a^2 x^2 - a^2 c^2 - a^2 y^2 + a^4 &= 0 \\c^2 y^2 - a^2 y^2 - a^2 x^2 &= a^2 c^2 - a^4 \\y^2 (c^2 - a^2) - x^2(a^2) &= a^2 (c^2 - a^2) \\\frac{y^2}{a^2} - \frac{x^2}{c^2 - a^2} &= 1

Like the collapse we witnessed when investigating the focal property for ellipses, this equation became pretty simple. We can even make it simpler. For any positive number b and a, there exists a c such that b^2 = c^2 - a^2. Thus the general form for a hyperbola that open upwards and downwards and whose foci lie on the y-axis is:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

Switching x and y we have hyperbolas that open rightwards and leftwards and whose foci lie on the x-axis.

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

These equations have so far been hyperbolas that are centered about the origin. For a hyperbola that is centered around the point (h,k) we have the shifted equations:

\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1

for a hyperbola opening up and down, and

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

For a hyperbola opening left and right.

Example 1

Show that the following equation is a hyperbola. Graph it, and show its foci.

144x^2 - 576x - 25y^2 - 150y - 3249 = 0

Solution: The positive leading coefficient for the x^2 term and the negative leading coefficient for the y^2 term indicate that this is a hyperbola that is horizontally oriented. Grouping and completing the square, we have:

144(x^2-4x)-25(y^2+6y)& = 3249 \\144(x^2 - 4x+4) - 25(y^2+6y+9)& = 3249 + 576 -225 \\144(x-2)^2-25(y+3)^2 & = 3600 \\\frac{(x-2)^2}{5^2} - \frac{(y+3)^2}{12^2} & = 1

So our hyperbola is centered at (2,-3). Its vertices are 5 units to the right and left of (2,-3), or at the points (7,-3) and (-3,-3). It opens to the right and left from these vertices. It’s foci are c units to the left and right of (2,-3), where c = \sqrt{a^2 + b^2} = \sqrt{5^2 + 12^2} = 13. So it’s foci are at (15,-3) and (-11,-3). Plotting a few points near (7,-3) and (-3,-3), the graph looks like:

Review Questions

  1. Explain why for any positive number b and a, there exists a c such that b^2=c^2-a^2.
  2. Graph the following hyperbola and mark its foci: 16x^2 + 64x-9y^2+90y - 305=0
  3. Graph the following hyperbola and mark its foci: 9y^2+18y-x^2+4x-4=0
  4. Graph the following hyperbola and mark its foci: 25x^2+150x-4y^2+24y+89=0
  5. Find the equation for the following hyperbola:

Review Answers

  1. Let c = \sqrt{a^2+b^2}. Since a^2 + b^2 is always positive for positive a and b, this number is always defined. Geometrically, let c be the hypotenuse of a right triangle with side lengths a and b.
  2. \frac{(x - 4)^2}{4} - \frac{(y + 2)^2}{45} = 1

Asymptotes

In addition to their focal property, hyperbolas also have another interesting geometric property. Unlike a parabola, a hyperbola becomes infinitesimally close to a certain line as the x- or y-coordinates approach infinity. Such a line is called an asymptote.

Before we try to prove this property of hyperbolas, we have to figure out what we mean by “infinitesimally close.” Here we mean two things: 1) The further you go along the curve, the closer you get to the asymptote, and 2) If you name a distance, no matter how small, eventually the curve will be that close to the asymptote. Or, using the language of limits, as we go further from the vertex of the hyperbola the limit of the distance between the hyperbola and the asymptote is 0.

So now we need to prove such a line (or lines) exist for the hyperbola. It turns out there are two of them, and that they cross at the point at which the hyperbola is centered:

It also turns out that for a hyperbola of the form \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the asymptotes are the lines y = \frac{b}{a} x and y = - \frac{b}{a} x. (for a hyperbola of the form \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 the asymptotes are the lines y = \frac{a}{b} x and y = - \frac{a}{b} x. For a shifted hyperbola, the asymptotes shift accordingly.)

To prove that the hyperbola actually gets infinitesimally close to these lines as it goes to infinity, let’s focus on the point P on the upper right leg of the hyperbola in the picture below.

Now consider the point P', the point on the asymptote lying directly above P. If we can show that P and P' become infinitesimally close, then that’s good enough (this subtlety will be explored in the next exercise.) Well, the point P is on the hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, so if it’s x-value is x, it’s y- value can be found by solving for y in the equation. So

\frac{x^2}{a^2} - 1 & = \frac{y^2}{b^2}\\y^2 & = \frac{b^2}{a^2} x^2 - b^2 \\y & = \sqrt{\frac{b^2}{a^2} x^2 - b^2}

This expression simplifies further, but we’ll leave it for now and work on P'. Since P' lies directly above P, x is also the x- coordinate of P'. Similarly as with P, we can find the y-coordinate from the equation of the asymptote. Since y = \frac{b}{a} x is the equation of the asymptote, we have \frac{b}{a} x as the y-coordinate of P'. So P = \left (x, \sqrt{ \frac {b^2}{a^2} x^2 - b^2} \right ), and P' = \left (x, \frac{b}{a} x \right ). Since these two points are vertically aligned, the distance between them points is simply the difference between the y-coordinate:

Distance between P and P' = \frac{b}{a} x - \sqrt{\frac{b^2}{a^2} x^2 - b^2}

& = \frac{b}{a} x - \sqrt{\frac{b^2}{a^2} x^2 - \frac{b^2}{a^2} a^2} \\& = \frac{b}{a} x - \sqrt{ \frac{b^2}{a^2}(x^2 - a^2)} \\& = \frac{b}{a} x - \frac{b}{a} \sqrt{(x^2 - a^2)} \\& = \frac{b}{a} \left (x - \sqrt{(x^2 - a^2)} \right )

So the distance is a constant multiplied by a quantity that depends on x. What happens to this quantity when x approaches infinity? This kind of question involves working with limits in a way you will study soon. For now, we can use a nice geometric picture to see what happens. In the picture below, we see that a, x, and \sqrt{(x^2 - a^2)} can be thought of as side-lengths of a right triangle, where x is the hypotenuse. This can be verified with the Pythagorean Theorem.

In this picture, what happens to difference in lengths between the side of length x and the side of length \sqrt{(x^2 - a^2)} when x approaches infinity and a remains fixed? The two side lengths become closer in length, so their difference approaches zero.

Going back to the previous diagram where we defined P and P', this means that P and P' become infinitesimally close as x approaches infinity.

This same argument can be repeated to show that the other three legs of the hyperbola approach their respective asymptotes, and to show that the same holds in vertically-oriented and/or shifted parabolas.

Example: Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: 9x^2 - 36x - 4y^2 - 16y - 16 = 0.

Solution: First, we put the hyperbola into the standard form:

9(x^2 - 4x) - 4(y^2 + 4y) & = 16 \\9(x^2 - 4x + 4) - 4(y^2 + 4y + 4) & = 36 \\\frac{(x - 2)^2}{4} - \frac{(y + 2)^2}{9} & = 1

So a = 2, b =3 and c = \sqrt{4 + 9}=\sqrt{13}. The hyperbola is horizontally oriented, centered at the point (2,-2), with foci at \left (2 + \sqrt{13}, -2 \right ) and \left (2 - \sqrt{13}, -2 \right ). After taking shifting into consideration, the asymptotes are the lines: y + 2 = \frac{3}{2} (x - 2) and y + 2 = - \frac{3}{2} (x - 2). So graphing the vertices and a few points on either side, we see the hyperbola looks something like this:

Review Questions

  1. In the diagram above, which was used in the proof that hyperbolas have asymptotes, why is it true that if P and P' become infinitesimally close, that this implies the hyperbola and the line become infinitesimally close. (HINT: what exactly do we mean by the “closeness” of a point on a hyperbola and a line?)
  2. Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: 16x^2 - 96x - 9y^2 - 36y - 84 = 0.
  3. Graph the following hyperbola, drawing its foci and asymptotes and using them to create a better drawing: y^2 - 14y -25x^2 - 200x -376 = 0.
  4. Find the equation for a hyperbola with asymptotes of slopes \frac{5}{12} and - \frac{5}{12}, and foci at points (2,11) and (2,1).
  5. A hyperbolas with perpendicular asymptotes is called perpendicular. What does the equation of a perpendicular hyperbola look like?

Review Answers

  1. The distance between a point and a line is the shortest segment between the point and a point on the line. We have shown that some distance—not necessarily the shortest—between P and a point on the asymptote becomes infinitesimally smaller. This means that the shortest distance between P and the asymptote must also become shorter.
  2. \frac{(y - 6)^2}{25} - \frac{(x - 2)^2}{144} = 1
  3. The slopes of perpendicular lines are negative reciprocals of each other. This means that \frac{a}{b} = \frac{b}{a}, which, for positive a and b means a = b.

Applications

Vexing Questions from Ancient Greece

This application of hyperbolas is the reason Menaechmus, one of the first serious thinkers about conic sections, got into them in the first place.

It is thought that conic sections first attracted the attention of the Ancient Greeks when Menaechmus used two conics to find a solution to another famous Greek problem. A famous class of Greek problems involves constructing a length with the simplest tools possible. One of these problems is often referred to as the problem of “doubling a cube.” The problem is to make a cube that is twice the volume of a given cube.

This, of course, does not simply mean making a cube whose sides are double in length. This would produce a cube with eight times the volume. If the first cube has side length one unit then it has volume 1^3 = 1. In order for a cube to have double that volume, 2, it must have a side length of \sqrt[3]{2}. So the problem boils down to producing a segment whose length is \sqrt[3]{2}. Once this is done, the cube can be constructed out of segments of this length.

So Menaechmus was working on constructing a segment of length \sqrt[3]{2}. Now you are probably thinking, “Easy, just plug \sqrt[3]{2} into your calculator.” But Menaechmus didn't have a calculator into which he could type 2^\wedge (1/3). An even more important point is that even if he did have such a calculator, this wouldn't solve the problem in a way Menaechmus would have been pleased with. He was trying to use the simplest technology possible, and a calculator—even if he had one—wouldn’t fit this bill. Plus, a calculator merely finds an estimate of this number, and the Greeks aimed to find a mathematical situation in which the exact number would be generated.

The very simplest tools the Greeks considered for making these kinds of constructions were an idealized compass and straightedge. Unknown to the Greeks at the time, a compass and straightedge don’t do the job for this length, though they didn’t know it at the time—it wasn’t discovered until 1837, when Pierre Laurent Wantzel demonstrated that this construction is impossible.

According to some historians (Heath [footnote]), Menaechmus did find a solution to the problem and he used conic sections to do so. Before his time, the problem had already been reduced to finding a specific set of “mean proportionals”, or a chain of numbers with equivalent subsequent ratios. In particular, to find numbers x and y such that 1 is to x as x is to y as y is to 2, or in fraction form:

\frac{1}{x} = \frac{x}{y} = \frac{y}{2}

How does this chain of proportions help find the number \sqrt[3]{2}? Well, looking at the first equality and cross-multiplying, we have y = x^2. Looking at the second equality, we have y^2 = 2x. Using the first equation to substitute x^2 for y in the second equation, we have (x^2)^2 = 2x, which reduces to x^3 = 2, the solution of which is \sqrt[3]{2}.

But solving the above chain of ratios is also equivalent to finding the intersection of the parabola y= x^2 and the hyperbola xy = 2. You’ll examine why this is true in an upcoming review questions.

But wait! Hold the presses! Did I say the hyperbola xy = 2 ? But that doesn’t look like the form of any of the hyperbolas we have looked at! This is because it is oriented with its foci on a diagonal, rather than horizontally or vertically. In the next section we will look at how rotating conic sections from their standard positions affects their forms. For now, you can check that xy = 2 indeed looks like a hyperbola by plugging in a few points near the origin—which you will do in the exercises below.

Review Questions

  1. Draw a sketch of xy = 2 by plotting a few points near the origin.
  2. What are the asymptotes of the hyperbola xy = 2 ? What are the foci?
  3. Explain why solving the mean proportional \frac{1}{x} = \frac{x}{y} = \frac{y}{2} is equivalent to finding the intersection of the parabola y = x^2 and the hyperbola xy = 2.
  4. In the language of “compass and straightedge,” though Menaechmus’ construction can’t be done with a traditional compass, it can be done with the generalized compass discussed in the previous section. Explain how this tool could be used to “double” an existing cube.

Review Answers

  1. The asymptotes are the x- and y-axes. The foci are (2,2) and (-2,-2) (these are relatively hard to find, but it is relatively easy to show they are the foci once they are found.)
  2. These two equations are obtained by looking at the first equality and cross multiplying, as well as setting the first term equal to the third term and cross multiplying. These two equalities hold exactly the same amount of information as the chain of equalities.
  3. Using a compass and straightedge, a coordinate grid with unit lengths can be drawn. See http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions for more on using a compass and straightedge to make geometric constructions. Upon this, the two shapes can be drawn using the generalized compass as discussed in the last session. The distance between the y-axis and the intersection point is of length \sqrt[3]{2}.

Vocabulary

Hyperbola
A conic section where the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shapes curves.
Unbounded
A shape which is so large that no circle, no matter how large, can enclose the shape.
Asymptote
A line which a curve approaches as the curve and the line approach infinity, eventually becoming closer than any given positive number.
Perpendicular Hyperbola
A hyperbola where the asymptotes are perpendicular.

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Feb 23, 2012

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Sep 24, 2014
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CK.MAT.ENG.SE.1.Math-Analysis.6.4

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