Conic Sections

This chapter explains the generation and properties of three important classes of shapes. It begins with an overview and then continues with individual sections on ellipses, parabolas, and hyperbolas.

19.1   The Standard Conic Sections

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The conic sections are shapes that appear when a double cone is intersected by a flat plane. The standard conic sections are circles, ellipses, parabolas, and hyperbolas, as shown in Fig. 19-1. Note that a parabola only occurs if the intersecting plane is parallel to the side of the cone. Also note that only the hyperbola intersects both the top and bottom halves of the cone.

Fig. 19-1

19.1A   Degenerate Conics

If the intersecting plane goes through the vertex of the cone, we get degenerate conics: straight lines, individual points, and pairs of lines, as shown in Fig. 19-2. Note that we only get a single line if the intersecting plane is parallel to the side of the cone.

Fig. 19-2

19.1B Conics as Loci of Equations

Any standard or degenerate conic section shape occurs as the locus of points satisfying an equation of the form ax² + by² + cx + dy + e = 0 for some constants a, b, c, d, and e. We can use the algebraic technique known as “completing the square” to rearrange such an equation into one of the following standard forms:

The equation of a circle with center (i, j) and radius r is (x − i)² + (y − j)² = r².

The equation of an ellipse is .

The equation of a parabola is either (y − j) = k(x − i)² or (x − i) = k(y − j)².

The equation of a hyperbola is .

The equation for a single point at (i, j) is (x − i)² + (y − j)² = 0.

The equation for the straight line through (i, j) with slope m is (y − j) = m(x − i).

The equation for a pair of lines that intersect at (i, j) is (y − j)² = m² (x − i)². If we take the square root of both sides, this results in two equations: (y − j) = ± m(x − i).

An equation of degree two in variables x and y is any equation of the form ax² + fxy + by² + cx + dy + e = 0 for constants a, b, c, d, e, and f. We can get every possible conic section shape with f = 0. For each f ≠ 0, we get a rotated version of one of these shapes. For example, the equation of a parabola whose axis is neither horizontal nor vertical will have an xy term. The algebra involved in working with an xy term is beyond the scope of this book, so this text will be limited to the standard forms described above, where f = 0.

19.1C   Conics from Focus Points and Directrix Lines

A third way to construct a conic section (but not a degenerate or a circle) is with a point F called a focus, a line l called a directrix, and a number e > 0 called the eccentricity. The locus of all points P for which PF = e × Pl will be an ellipse if 0 < e < 1, a parabola if e = 1, and a hyperbola if e > 1. That is, a parabola consists of all points P whose distance PF to the focus is exactly the same as its distance Pl to the directrix. An ellipse consists of all points that are closer to F than to l (by exactly the factor e), and a hyperbola consists of all points that are further from F than from l (further by a multiple of e > 1). See Fig. 19-3.

Fig. 19-3

Note that when e > 1, there are points on both sides of the directrix that are the right amount further from the focus than the directrix, and thus this method generates both pieces of the hyperbola. Because of their reflectional symmetries, hyperbolas and ellipses have two sets of foci and directrices. Parabolas, by contrast, have only one focus and one directrix.

19.1D   Reflection Properties

To illustrate this concept, we will imagine ellipses, parabolas, and hyperbolas made of a reflective material and trace the paths of light traveling to and from the focus points.

For ellipses, light traveling toward one focus will reflect away from the other focus. Light traveling from one focus will reflect toward the other.

For parabolas, light traveling from the focus will reflect so as to be parallel to the axis of reflection. Light parallel to this axis will reflect toward the focus on one side of the parabola and away from it on the other.

For hyperbolas, light traveling toward one focus will reflect toward the other, and light traveling away from a focus will reflect away from the other.

These properties are illustrated in the sections that follow, along with real-life applications. The proofs require calculus, so they are omitted.

19.2   Ellipses

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An ellipse looks like a stretched circle. For example, if we dihilate a unit circle by a factor of 2 in just the x direction using the transformation T(x, y) T’ (2x, y), we get an ellipse, as shown in Fig. 19-4.

Fig. 19-4

19.2A   Parts of an Ellipse

An ellipse has a center point around which there is a 180º rotational symmetry. The longest line segment through the center is called the major axis. Perpendicular to this, also through the center, is the minor axis (the shortest distance across the ellipse). These axes are the two axes of symmetry for the ellipse. The endpoints of the axes are called vertices. Along the major axis will be the two foci, with their directrices on the outside.

Fig. 19-5

In Fig. 19-5, the center is O, the major axis is , and the minor axis is . The points A, B, C, and D are the vertices. Point F₁ is the focus that corresponds with directrix l₁, and F₂ is the focus that corresponds with l₂. Note that if we used F₁ with l₂, we would get a different ellipse. Also note that if were longer than , the two foci would be on and the two directrices would be horizontal.

19.2B   Equations of Ellipses

To find the equation of an ellipse, let (i, j) be the coordinates of the center, let A be half the length of the horizontal axis, and let B be half the length of the vertical axis. The ellipse is the locus of the equation . Furthermore, the two foci are found by moving a distance of away from the center in both directions along the major axis.

SOLVED PROBLEMS: FINDING EQUATIONS OF ELLIPSES

19.1   Find the equations and foci of the ellipses in Fig. 19-6.

Fig. 19-6

Solutions

(a) The center of this ellipse is (−3, 0), so i = −3 and j = 0. The horizontal axis is 2 units long, so A = 1. The vertical axis is 6 units tall, so B = 3. Here, the major axis is the vertical one. The equation is . To find the foci, we travel units up and down the major axis from the center to get (−3, 2.83) and (−3, −2.83).

(b) The center of this ellipse is the point (3, −2), so i = 3 and j = −2. The horizontal axis has length 6 and the vertical axis has length 4, so A = 3 and B = 2. Thus the equation of this ellipse is . The major axis is horizontal, so we find the two foci by traveling a distance of to the left and right of the center, obtaining (5.24, −2) and (0.76, 2).

SOLVED PROBLEMS: SKETCHING ELLIPSES

19.2   Sketch the graph of each ellipse and plot their foci.

(b)  9x² + 25(y − 2)² = 225

Solutions

(a) Here, the center is O = (−1, 1), the horizontal axis is 2(3) = 6 units long, and the vertical axis is 2(4) = 8 units long. After plotting the vertices 3 units to the left and right of the center, and 4 units above and below, we can then sketch a rounded shape whose curves turn around at these points. The foci are units from the center. Because the major axis is vertical here, the foci are above and below the center at F₁ = (−1, 3.65) and F₂ = (−1, −1.65). This is shown in Fig. 19-7(a).

(b) To get this equation into standard form, we must divide both sides by 225 and then represent the denominators as squares: . This ellipse has its center at (0, 2) and vertices at (0 ± 5,2) and (0,2 ± 3). That is, the vertices are (5, 2), (−5, 2), (0, −1), and (0, 5). The two foci are found by moving units from the center along the horizontal major axis; thus, F₁ = (−4, 2) and F₂ = (4, 2), as shown in Fig. 19-7(b).

Fig. 19-7

19.2C   Elliptical Reflections

Now suppose that we have an ellipse made of a reflective material. Light traveling toward a focus from the outside of the ellipse will bounce directly away from the second focus. In Fig. 19-8(a), for example, the ray of light aimed directly at focus F₁ reflects straight away from focus F₂.

Similarly, light traveling from one focus will always reflect directly toward the other focus. This is shown in Fig. 19-8(b). Furthermore, when light travels from one focus to the other inside an ellipse like this, the sum of the distances F₁P + F₂P will always be the length of the major axis.

Fig. 19-8

Now suppose that two people stand at the foci of a large, elliptical room. A whisper from one person will travel out in all directions, then bounce off the walls directly toward the second person. Each bit of whisper will travel the same distance, so the sounds will all arrive at the same time and be audible.

19.3   Parabolas

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A parabola is an infinite curve with a single axis of reflection. On this axis, there is the focus point F, a single vertex point V, and the point of intersection with the directrix. See Fig. 19-9.

Fig. 19-9

19.3A   Equations of Parabolas

The standard equation for a parabola whose vertex is at (i, j) is either

(y − j) = k(x − i)² (opening up if k > 0 and down if k < 0) or else

(x − i) = k(y − j)² (opening to the right if k > 0, and to the left if k < 0).

If we step 1 unit perpendicular to the axis, a point of the parabola will be |k| units in the direction of the axis. This can be used to sketch parabolas and identify their equations.

The focus of a parabola is found by traveling units along the axis from the vertex.

SOLVED PROBLEMS: FINDING EQUATIONS OF PARABOLAS

19.3   Find the equation and focus for each of the parabolas in Fig. 19-10.

Fig. 19-10

Solutions

(a) The vertex is (i, j) = (0, 3) and the parabola opens upward, so the equation has the form (y − 3) = k(x − 0)² for a positive k. If we take a step to the right from the vertex at x = 0, we have to move up k = 2 units to the point (1, 5) on the parabola. Thus, the equation is y − 3 = 2x², although many would write this as y = 2x² + 3. Since k = 2, the focus is of a unit up from the vertex, at (0, 3.125).

(b) The vertex is (i, j) = (−3, 1) and the parabola opens downward, so we have (y − 1) = k(x − (−3))² with a negative k. If we move 1 unit to the right of the vertex, the next point is down 1 unit at (−2, 0). Thus k = − 1 and the equation is y − 1 = − (x + 3)². To find the focus, we compute and go 1/4 of a unit below the vertex to (−3, 0.75).

(c) Here, the vertex is (i, j) = (3, 2) and the parabola opens to the right, so our equation has the form (x − 3) = k(y − 2)² for a positive k. If we step down 1 unit from the vertex, it appears as though the next point is k = 1/2 of a unit over at (3.5, 1). Thus, the equation is (x − 3) = (y − 2)². Precision is difficult when reading points on a graph, so it is possible that k is really 0.49 or 0.52 or something else close to 0.5. Assuming k = 1/2, the focus is a unit to the right of the vertex, at (3.5, 2).

The manner in which we obtain the values of k is shown in Fig. 19-11.

Fig. 19-11

SOLVED PROBLEMS: SKETCHING PARABOLAS

19.4   Sketch the graphs of each parabola and find their foci:

(a)  x − 5 = −2(y + 2)²

(b)  y − 1 = (x + 3)²

Solutions

(a) The vertex is where x = 5 and y = −2, so (i, j) = (5, −2). As the y is squared, this parabola has a horizontal axis of symmetry. Because k = −2, it opens to the left, and when we move 1 unit up or down from the vertex, we must move 2 units to the left. The focus is units from the vertex along the horizontal axis, at (4.875, −2). This is sketched in Fig. 19-12(a).

(b) Here, the vertex is (i, j) = (−3, 1) and the parabola opens upward. If we move 1 unit to the left or right, we move up by k = 1/2. The focus is found by going of a unit along the axis from the vertex at (−3, 1.5), as shown in Fig. 19-12(b).

Fig. 19-12

19.3B   Parabolic Reflections

The reflective property of parabolas is shown in Fig. 19-13. Light beams coming from the focus will all reflect to beams parallel to the axis, as shown in (a). If we reverse the light, light parallel to the axis will reflect directly to the focus, as shown in (b). If light beams in the direction of the axis strike the convex side of the parabola, they will each scatter directly away from the focus, as shown in (c). If this is reversed, all beams headed toward the focus will reflect into parallel beams, as shown in (d).

Fig. 19-13

The property in (a) is used when a concave parabolic mirror is placed behind the bulb of a flashlight. If the light source is at the focus, the light will shine out in a straight, narrow beam.

The property in (b) is used when a satellite dish is shaped like a parabola. If the dish is pointed directly at the satellite, then the incoming transmission waves will be basically parallel and reflect toward a single focal point. A small receiver suspended at this point will thus receive a greatly magnified signal.

The property in (d) is used when a person installs a convex parabolic mirror across the street from a driveway. Everything that the focal point can “see” (have a beam of light travel straight to it) will appear on the surface of the mirror to someone in the direction of the axis. Similarly, a parabolic mirror in the corner of a large room will allow someone (in the direction of the axis) to see everything in the room at once. If that person were to shine a light at the mirror, everyone in the room would see the spot of light as if it were at the mirror’s focal point, because of (c).

19.4   Hyperbolas

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A hyperbola consists of two disconnected curves, shown with solid lines in Fig. 19-14. There is 180º rotational symmetry around the center point O, and two axes of symmetry. The transverse axis T runs through the two foci (F₁ and F₂) and the two vertices (V₁ and V₂). The conjugate axis C is perpendicular to the transverse axis at the center point and contains no points of the hyperbola. The two directrix lines l₁ and l₂ are parallel to the conjugate axis. Finally, as the points of the hyperbola get further from the center, they get closer to two slant asymptotes S₁ and S₂.

Fig. 19-14

19.4A   Equations of Hyperbolas

The standard form for the equation of a hyperbola is

(a)   if the hyperbola opens to the left and right, and

(b)   if the hyperbola opens up and down.

In both cases, the center is (i, j). If we draw a rectangle that is 2A units wide, 2B units tall, and centered at (i, j), then the diagonals of the rectangle will extend out to be the slant asymptotes.

In (a), the conjugate axis is x = i (because if x = i, the equation can’t be solved) and the transverse axis is y = j. In (b), this is reversed: the conjugate axis is y = j and the transverse axis is x = i.

The two vertices are where the transverse axis crosses the rectangle we have drawn. The two foci are the points that are units from the centere along the transverse axis.

SOLVED PROBLEMS: SKETCHING HYPERBOLAS

19.5   Sketch the graphs and find the foci of:

Solutions

(a) We have center (3, −1), conjugate axis x = 3, and transverse axis y = −1. As A = 4 and B = 2, we draw an 8 by 4 rectangle around the center and extend out its diagonals for the slant asymptotes. Since the transverse axis is horizontal, the midpoints of the two sides of the rectangle are the vertices: (−1, −1) and (7, 1). By drawing curving lines from these vertices out, away from the center, and approaching the slant asymptotes, we get a decent approximation of the curve, as shown in Fig. 19-15(a). To obtain the foci, we compute and travel this distance from the center along the transverse axis to just beyond the curve to F₁ = (−1.47, −1) and F₂ = (7.47, −1).

(b) The center is (0, 2) with A = 3 and B = 2. We draw a rectangle that is 6 units wide and 4 units tall around the center and extend out its diagonals for the slant asymptotes. Of the two axes, if we set x = 0, we get , whose two solutions y = 4 and y = 0 are the vertices. If we set y = 2, however, we get , which is impossible; thus y = 2 is the conjugate axis. As before, we plot the two vertices and draw curves that approach the asymptotes. The foci are then units along the transverse axis from the center, at F₁ = (0, −1.6) and F₂ = (0, 5.6), as shown in Fig. 19-15(b).

Fig. 19-15

SOLVED PROBLEMS: FINDING EQUATIONS OF HYPERBOLAS

19.6   Estimate the equations of the hyperbolas in Fig. 19-16.

Fig. 19-16

Solutions

(a) It is clear that the two vertices are at (4, 4) and (4, −2), and thus the center is the midpoint of these, (4, 1) = (i, j). Furthermore, since the vertices are each 3 vertical units away from the center, this means that B = 3 (it would have been A if the transverse axis were horizontal). Next, we sketch in slant asymptotes through the center and connect them with horizontal lines through the vertices, as shown in Fig. 19-17(a). We then complete the rectangle, as shown with broken vertical lines, and see that it measures 2 horizontally and 6 vertically. This means that A = 1, B = 3, and the equation must be (we know the x² term must be negative, because the conjugate axis is horizontal).

(b) Here, the vertices are (−3, 1) and (−1, 1), so the center is (i, j) = (−2, 1). To sketch the slant asymptotes, we go an equal distance up and down the hyperbola from the center, say 9 units, and trace lines through the center. If we draw vertical lines through the vertices until they meet these slant lines and then connect them with horizontal, broken lines, we get the rectangle in Fig. 19-17(b). Clearly, the horizontal width is 2A = 2 and the vertical height of the rectangle is 2B = 4, so A = 1 and B = 2. This time, the y² term will be negative, so our equation is or just .

Fig. 19-17

19.4B   Hyperbolic Reflections

If a hyperbola is made of reflective material, then light coming away from one focus will reflect away from the other focus and light going toward one focus will reflect toward the other focus. For example, in Fig. 19-18, light going away from F₁ reflects so that it ends up traveling directly away from F₂ in both situations (a) and (b). Similarly, light going toward F₁ reflects toward F₂ in both case (c) and case (d).

Fig. 19-18

If we have a mirror shaped like half of a hyperbola and take a picture of it from the outside focus (Fig. 19-19), the result will be a panoramic photo. Every illuminated object within sight of the mirror will send some light toward the inner focus F₁, and all of this light will be reflected to the camera. The photograph of the mirror will thus capture all the light rays going toward F₁, and thus look exactly like the view from that location: a panorama.

Fig. 19-19

SUPPLEMENTARY PROBLEMS

19.1. For each of the following ellipses, find the center, both axes, all vertices, and both foci.

19.2. For each of the following parabolas, find the vertex, the axis of symmetry, and the focus:.

(a)  (x − 4) = 3(y + 2)²

(b)  y = −0.1x² − 2

(c)  (x + 2) = (y + 1)²

19.3. For each of the following hyperbolas, find the center, axes, vertices, and foci.

19.4. Sketch the graph of each of the following equations, including focus point(s):

19.5. Find the equation (in standard form) for each of the graphs in Fig. 19-20.

Fig. 19-20

19.6. Suppose that the curve in Fig. 19-21 is a reflective parabola with focus F. Suppose further that the rays at (a) and (d) are parallel to the axis of symmetry, and that (b) is traveling toward F. Illustrate the directions in which each of these rays will bounce.

Fig. 19-21

19.7. Suppose that the curve in Fig. 19-22 is a reflective ellipse with foci F₁ and F₂, and that the rays in (c) and (d) are traveling toward foci F₂ and F_1, respectively. Illustrate the directions in which each of these rays will bounce.

Fig. 19-22

19.8. Suppose that the curve in Fig. 19-23 is a reflective hyperbola with foci F₁ and F₂. Suppose further that each ray is either traveling directly toward or directly away from one of these foci. Illustrate the directions in which each of these rays will bounce.

Fig. 19-23