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Here is a list of the most important curves in mathematics, so you can tell what is coming. It is not easy to rank the top four:

**straight lines****sines and cosines**(oscillation)**exponentials**(growth and decay)**parabolas, ellipses, and hyperbolas**(using 1, x, y, x², xy, y²).

The curves that I wrote last, the Greeks would have written first. It is so natural to go from linear equations to quadratic equations. Straight lines use 1, x, y. Second degree curves include x², xy, y². If we go on to x³ and y³, the mathematics gets complicated. We now study equations of second degree, and the curves they produce.

It is quite important to see both the **equations** and the **curves**. This section connects two great parts of mathematics—analysis of the equation and geometry of the curve. Together they produce "**analytic geometry**." You already know about functions and graphs. Even more basic: Numbers correspond to points. We speak about "the point (5,2)." Euclid might not have understood.

[Fig. 3.15]

Where Euclid drew a 45° line through the origin, Descartes wrote down y = x. Analytic geometry has become central to mathematics-we now look at one part of it.

The parabola and ellipse and hyperbola have absolutely remarkable properties. The Greeks discovered that all these curves come from **slicing a cone by a plane**. The curves are "conic sections." A level cut gives a **circle**, and a moderate angle produces an **ellipse**. A steep cut gives the two pieces of a **hyperbola** (Figure 3.15d). At the borderline, when the slicing angle matches the cone angle, the plane carves out a **parabola**. It has one branch like an ellipse, but it opens to infinity like a hyperbola.

Throughout mathematics, parabolas are on the border between ellipses and hyperbolas.

To repeat: We can slice through cones or we can look for equations. For a cone of light, we see an ellipse on the wall. (The wall cuts into the light cone.) For an equation AX² + Bxy + Cy² + Dx + Ey + F = 0, we will work to make it simpler. The graph will be centered and rescaled (and rotated if necessary), aiming for an equation like y = x². Eccentricity and polar coordinates are left for Chapter 9.

You knew this function long before calculus. The graph crosses the x axis when y = 0. The quadratic formula solves y = 3x² − 4x + 1 = 0, and so does factoring into (x − 1)(3x − 1). The crossing points x = 1 and x = 1/3 come from algebra.

The other important point is found by calculus. It is the minimum point, where dy/dx = 6x − 4 = 0. The x coordinate is 4/6 = 2/3, halfway between the crossing points. The height is y_{min} = -1/3. This is the vertex V in Figure 3.16a—at the bottom of the parabola.

A parabola has no asymptotes. The slope 6x − 4 doesn't approach a constant.

To center the vertex Shift left by 2/3 and up by 1/3. So introduce the new variables X = x − 2/3 and Y = y + 1/3. hen x = 2/3 and y= -1/3 correspond to X = Y = 0 —which is the new vertex:

y = 3x² − 4x + 1 becomes Y = 3X². (1)

Check the algebra. Y = 3X² is the same as y + 1/3 = 3(x − 3)². That simplifies to the original equation y = 3x² − 4x + 1. The second graph shows the centered parabola Y = 3X², with the vertex moved to the origin.

*To zoom in on the vertex* Rescale X and Y by the zoom factor a:

Y = 3x² becomes y/a = 3(x/a)².

The final equation has x and y in boldface. With a = 3 we find y = x² —the graph is magnified by 3. In two steps we have reached the model parabola opening upward.

[Fig. 3.16]

A parabola has another important point-the focus. Its distance from the vertex is called p. The special parabola y = x² has p = 114, and other parabolas Y = ax² have p = 1/4a. You magnify by a factor a to get y = x². The beautiful property of a parabola is that **every ray coming straight down is reflected to the focus**.

Problem 2.3.25 located the focus F—here we mention two applications. A solar collector and a TV dish are parabolic. They concentrate sun rays and TV signals onto a point—a heat cell or a receiver collects them at the focus. The 1982 UMAP Journal explains how radar and sonar use the same idea. Car headlights turn the idea around, and send the light outward.

Here is a classical fact about parabolas. **From each point on the curve, the distance to the focus equals the distance to the "directrix."** The directrix is the line y = -p below the vertex (so the vertex is halfway between focus and directrix). With p = 4, the distance down from any (x, y) is y + 1/4. Match that with the distance to the focus at (0,1/4)—this is the square root below. Out comes the special parabola y = x²:

y + 1/4 = √x² + (y − 1/2)²— (square both sides) — y = x². (2)

The exercises give practice with all the steps we have taken-center the parabola to Y = ax², rescale it to y = x², locate the vertex and focus and directrix.

*Summary for other parabolas* y = ax² + bx + c has its vertex where dy/dx is zero. Thus 2ax + b = 0 and x = -b/2a. Shifting across to that point is "completing the square":

ax² + bx + c equals a(x + b/2a)² + C. (3)

Here C = c -(b2/4a) is the height of the vertex. The centering transform X = x + (b/2a), Y = y − C produces Y = ax². It moves the vertex to (0, 0), where it belongs.

For the ellipse and hyperbola, our plan of attack is the same:

- Center the curve to remove any linear terms Dx and Ey.
- Locate each focus and discover the reflection property.
- Rotate to remove Bxy if the equation contains it.

This equation makes the ellipse symmetric about (0, 0)—the center. Changing x to -x or y to -y leaves the same equation. No extra centering or rotation is needed.

The equation also shows that x²/a² and y²/b² cannot exceed one. (They add to one and can't be negative.) Therefore x² ≤ a², and x stays between -a and a. Similarly y stays between b and -b. The ellipse is inside a rectangle.

By solving for y we get a function (or two functions!) of x:

y²/b² = 1 − x²/a² gives y/b = ±√1 − x²/a² or y = ± (b/a)√a² − x².

The graphs are the top half (+) and bottom half (-) of the ellipse. To draw the ellipse, plot them together. They meet when y = 0, at x = a on the far right of Figure 3.17 and at x = -a on the far left. The maximum y = b and minimum y = -b are at the top and bottom of the ellipse, where we bump into the enclosing rectangle.

A circle is a special case of an ellipse, when a = b. The circle equation x² + y² = r² is the ellipse equation with a = b = r. This circle is centered at (0,0); other circles are centered at x = h, y = k. The circle is determined by its radius r and its center (h, k):

**Equation of circle**: (x − h)² + (y − k)²= r². (4)

In words, the distance from (x, y) on the circle to (h, k) at the center is r. The equation has linear terms -2hx and -2ky —they disappear when the center is (0,0).

*EXAMPLE 1* Find the circle that has a diameter from (1,7) to (5, 7).

Solution The center is halfway at (3,7). So r = 2 and (x − 3)² + (y − 7)² = 22.

*EXAMPLE2* Find the center and radius of the circle x2 -6x + y2 -14y = -54.

Solution Complete x² − 6x to the square (x − 3)² by adding 9. Complete y² − 14y to (y − 7)² by adding 49. Adding 9 and 49 to both sides of the equation leaves (x − 3)² + (y − 7)² = 4 —the same circle as in Example 1.

Quicker Solution Match the given equation with (4). Then h = 3, k = 7, and r = 2:

x² − 6x + y² -14y = -54 must agree with x² − 2hx + h² + y² − 2ky + k² = r².

The change to X = x − h and Y= y − k moves the center of the circle from (h, k) to (0,0). This is equally true for an ellipse:

The ellipse | (x − h)² | + | (y − k)² | = 1 becomes | X² | + | Y² | = 1 |

a² | b² | a² | b² |

When we rescale by x = X/a and y = Y/b, we get the unit circle x² + y² = 1.

The unit circle has area n. The ellipse has area nab (proved later in the book). The distance around the circle is 2π. The distance around an ellipse does not rescale—it has no simple formula.

[Fig. 3.17]

Now we leave circles and concentrate on ellipses. They have *two foci* (pronounced fo-sigh). For a parabola, the second focus is at infinity. For a circle, both foci are at the center. The foci of an ellipse are on its longer axis (its major axis), one focus on each side of the center:

F₁ is at x = c = √a² − b² and F₂ is at x = -c.

The right triangle in Figure 3.17 has sides a, b, c. From the top of the ellipse, the distance to each focus is a. From the endpoint at x = a, the distances to the foci are a + c and a − c. Adding (a + c) + (a − c) gives 2a. As you go around the ellipse, the distance to F₁ plus the distance to F₂ is constant (always 2a).

*3H* At all points on the ellipse, the sum of distances from the foci is 2a. This is another equation for the ellipse:

from F₁ and F₂ to (x, y): √(x − c)² + y² + √(x + c)² + y² = 2a. (5)

To draw an ellipse, tie a string of length 2a to the foci. Keep the string taut and your moving pencil will create the ellipse. This description uses a and c—the other form uses a and b (remember b² + c² = a²). Problem 24 asks you to simplify equation (5) until you reach x²/a² + y²/b² = 1.

The "whispering gallery" of the United States Senate is an ellipse. If you stand at one focus and speak quietly, you can be heard at the other focus (and nowhere else). Your voice is reflected off the walls to the other focus—following the path of the string. For a parabola the rays come in to the focus from infinity—where the second focus is.

A hospital uses this reflection property to split up kidney stones. The patient sits inside an ellipse with the kidney stone at one focus. At the other focus a lithotripter sends out hundreds of small shocks. You get a spinal anesthetic (I mean the patient) and the stones break into tiny pieces.

The most important focus is the Sun. The ellipse is the orbit of the Earth. See Section 12.4 for a terrible printing mistake by the Royal Mint, on England's last pound note. They put the Sun at the center.

*Question 1* Why do the whispers (and shock waves) arrive together at the second focus?

Answer Whichever way they go, the distance is 2a. Exception: straight path is 2c.

*Question 2* Locate the ellipse with equation 4x² + 9y² = 36.

Answer Divide by 36 to change the constant to 1. Now identify a and b:

x²/9 + y²/4 = 1 so a = √9 and b = √4. Foci at ± √9 − 4 = ±√5.

*Question 3* Shift the center of that ellipse across and down to x = 1, y = -5.

Answer Change x to x − 1. Change y to y + 5. The equation becomes (x − 1)²/9 + (y + 5)²/4 = 1. In practice we start with this uncentered ellipse and go the other way to center it.

**Notice the minus sign for a hyperbola.** That makes all the difference. Unlike an ellipse, x and y can both be large. The curve goes out to infinity. It is still symmetric, since x can change to -x and y to -y.

The center is at (0, 0). Solving for y again yields two functions (+ and -):

y²/a² − x²/b² = 1 gives y/a = ± √1 + x²/b² or y = √b² + x². (6)

The hyperbola has two branches that never meet. The upper branch, with a plus sign, has y ≥ a. The **vertex** V₁ is at x = 0, y = a —the lowest point on the branch. Much further out, when x is large, the hyperbola climbs up beside its **sloping asymptotes**:

if x²/b² = 1000 then y²/b² = 1001. So y/a is close to x/b or -b/x.

[Fig. 3.18]

The asymptotes are the lines y/a = x/b and y/a = -x/b. Their slopes are a/b and -a/b. You can't miss them in Figure 3.18.

For a hyperbola, the foci are inside the two branches. Their distance from the center is still called c. But now c = √a² + b², which is larger than a and b. The vertex is a distance c − a from one focus and c + a from the other. The diflerence (not the sum) is (c + a) − (c − a) = 2a.

All points on the hyperbola have this property: **The diflerence between distances to the foci is constantly 2a.** A ray coming in to one focus is reflected toward the other. The reflection is on the outside of the hyperbola, and the inside of the ellipse.

Here is an application to navigation. Radio signals leave two fixed transmitters at the same time. A ship receives the signals a millisecond apart. Where is the ship? Answer: It is on a hyperbola with foci at the transmitters. Radio signals travel 186 miles in a millisecond, so 186 = 2a. This determines the curve. In Long Range Navigation (LORAN) a third transmitter gives another hyperbola. Then the ship is located exactly.

*Question 4* How do hyperbolas differ from parabolas, far from the center?

Answer Hyperbolas have asymptotes. Parabolas don't.

The hyperbola has a natural rescaling. The appearance of x/b is a signal to change to X. Similarly y/a becomes Y. Then Y = 1 at the vertex, and we have a standard hyperbola:

y²/a² − x²/b² = 1 becomes Y² − X² = 1

A 90° turn gives X² − Y² = 1 —the hyperbola opens to the sides. A 45° turn produces 2XY = 1. We show below how to recognize x² + xy + y² = 1 as an ellipse and x² + 3xy + y² = 1 as a hyperbola. (They are not circles because of the xy term.) When the xy coefficient increases past 2, x² + y² no longer indicates an ellipse.

*Question 5* Locate the hyperbola with equation 9y² − 4x² = 36.

Answer Divide by 36. Then y²/4 − x²/9 = 1. Recognize a = √4 and b = √9.

*Question 6* Locate the uncentered hyperbola 9y² − 18y − 4x² − 4x = 28.

Answer Complete 9y² − 18y to 9(y − 1)² by adding 9. Complete 4x² + 4x to 4(x + ½)² by adding 4(½)² = 1. The equation is rewritten as 9(y − 1)² − 4(x + ½)² = 28 + 9 − 1. This is the hyperbola in Question 5 —except its center is (-½, 1).

To summarize: Find the center by completing squares. Then read off a and b.

This equation is of second degree, containing any and all of 1, x, y, x², xy, y². A plane is cutting through a cone. **Is the curve a parabola or ellipse or hyperbola?** Start with the most important case Ax² + Bxy + Cy² = 1.

*3l* The equation Ax² + Bxy + cy² = 1 produces a hyperbola if B² > 4AC and an ellipse if B² < 4AC. A parabola has B² = 4AC.

To recognize the curve, we remove Bxy by rotating the plane. This also changes A and C—but the combination B² − 4AC is not changed (proof omitted). An example is 2xy = 1, with B² = 4. It rotates to y² − x² = 1, with -4AC = 4. That positive number 4 signals a hyperbola-since A = -1 and C = 1 have opposite signs.

Another example is x² + y² = 1. It is a circle (a special ellipse). However we rotate, the equation stays the same. The combination B² − 4AC = 0 − 4⋅1⋅1 is negative, as predicted for ellipses.

To rotate by an angle a, change x and y to new variables x' and y':

x = x' cos α - y' sin α | and | x'= x cos α + y sin α | (7) |

y = x' sin α + y' cos α | y' = - y sin α + x cos α. |

Substituting for x and y changes Ax² + Bxy + cy² = 1 to A'x'² + B'x'y' + C'y'² = 1. The formulas for A', B', C' are painful so I go to the key point:

**B' is zero if the rotation angle a has tan 2α = B/(A − C).**

With B' = 0, the curve is easily recognized from A'x'² + C'y'² = 1. It is a hyperbola if A' and C' have opposite signs. Then B'² − 4A'C' is positive. The original B² − 4AC was also positive, because this special combination stays constant during rotation.

After the xy term is gone, we deal with x and y—by **centering**. To find the center, complete squares as in Questions 3 and 6. For total perfection, rescale to one of the model equations y = x² or x² + y² = 1 or y² − x² = 1.

The remaining question is about F = 0. What is the graph of Ax² + Bxy + cy² = O? The ellipse-hyperbola-parabola have disappeared. But if the Greeks were right, the cone is still cut by a plane. The degenerate case F = 0 occurs when the plane cuts **right through the sharp point of the cone**.

A level cut hits only that one point (0,0). The equation shrinks to x² + y² = 0, a circle with radius zero. A steep cut gives two lines. The hyperbola becomes y² − x² = 0, leaving only its asymptotes y = ±x. A cut at the exact angle of the cone gives only one line, as in x² = 0. **A single point, two lines, and one line** are very extreme cases of an ellipse, hyperbola, and parabola.

All these "conic sections" come from planes and cones. The beauty of the geometry, which Archimedes saw, is matched by the importance of the equations. Galileo discovered that projectiles go along parabolas (Chapter 12). Kepler discovered that the Earth travels on an ellipse (also Chapter 12). Finally Einstein discovered that light travels on hyperbolas. That is in four dimensions, and not in Chapter 12.

equation | vertices | foci | |
---|---|---|---|

P | y = ax² + bx + c | (-b/2a, c−b²/4a) | 1/4a above vertex, also infinity |

E | x²/a² + y²/b² = 1, a>b | (a, 0) and (-a, 0) | (c, 0) and (-c, 0): c = √a² − b² |

H | y²/b² − x²/a² = 1 | (0, a) and (0, -a) | (0, c) and (0, -c): c = √a² + b² |