Looking at parabolas with vertices at the origin

Parabolas can have graphs that go every which way and have vertices at any point in the coordinate system. When possible, though, you like to deal with parabolas that have their vertices at the origin. The equations are easier to deal with, and the applications are easier to solve. Therefore, I put the cart before the horse with this section to cover these specialized parabolas. (In the section “Observing the general form of parabola equations” later in the chapter, I deal with all the other parabolas you may run across.)

Opening to the right or left

Parabolas with their vertices (the plural form of vertex is vertices — just a little Latin for you) at the origin and opening to the right or left have a standard equation and are known as relations; you see a relationship between the variables. The standard form comes packed with information about the focus, directrix, vertex, axis of symmetry, and direction of a parabola. The equation also gives you a hint as to whether the parabola is narrow or opens wide.

The general form of a parabola with the equation gives you the following info:

• Focus: (a, 0)
• Directrix:
• Vertex: (0, 0)
• Axis of symmetry:
• Opening: To the right if a is positive; to the left if a is negative
• Shape: Narrow if   is less than 1; wide if   is greater than 1

I use the absolute value operation, , instead of saying that 4a has to be between 0 and 1 or between and 0. I think it’s just a neater way of dealing with the rule.

For example, if you want to extract information about the parabola , you can put it in the form by writing it . In this case, you extract the following info:

• The value of a is 2 (from ).
• The focus is at (2, 0).
• The directrix is the line .
• The vertex is at (0, 0).
• The axis of symmetry is .
• The parabola opens to the right.
• The parabola is wide because   is greater than 1.

Figure 11-3 shows the graph of the parabola with all the key information listed in the sketch.

Of course, extracting information is always more convenient when the equation of the parabola has an even-number coefficient, such as the 8 in , but the process works for odd numbers, too. You just have to deal with fractions. For instance, you can write the parabola as , so the value of a is , the focus is , and so on. The value of a is just what it takes to multiply 4 by to get the coefficient.

Opening upward or downward

Parabolas that open left or right are relations, but parabolas that have curves opening upward or downward are a bit more special. The parabolas that open upward or downward are functions — they have only one y value for every x value. (For more on functions, refer to Chapter 6.) Parabolas of this variety with their vertex at the origin have the following standard equation: .

You can tell that you have a relation (opening left or right) rather than a function (opening up or down) if the y variable is squared. The quadratic polynomials (from Chapter 7) and the functions all have the x variable squared.

The information you glean from the standard equation tells much the same story as the equation of parabolas that open sideways; however, many of the rules are reversed.

From the general form for the parabola , you extract the following information:

• Focus: (0, a)
• Directrix:
• Vertex: (0, 0)
• Axis of symmetry:
• Opening: Upward if a is positive; downward if a is negative
• Shape: Narrow if   is less than 1; wide if   is greater than 1

You convert the parabola to the form because the general form gives you quick, easily accessible information. To perform the conversion, just divide the coefficient by 4, and then write the coefficient as 4 times the result of the division. You haven’t changed the value of the coefficient; you’ve just changed how it looks.

In this case, you extract the following info:

• The value of a is .
• The focus falls on the point .
• The directrix forms a line at .
• The vertex is at (0, 0).
• The axis of symmetry is .
• The graph opens downward.
• The parabola is narrow because  .

Figure 11-4 shows the graph of the parabola with all the elements illustrated in the sketch.

Observing the general form of parabola equations

The curves of parabolas can open upward, downward, to the left, or to the right, but the curves don’t always have to have their vertices at the origin. Parabolas can wander all around a graph. So, how do you track the curves down to pin them on a graph? You look to their equations, which give you all the information you need to find out where they’ve wandered to.

When the vertex of a parabola is at the point (h, k), the general form for the equation is one of the following:

• Opening left or right: . When the y variable is squared, the parabola opens left or right. From this equation, just like with parabolas that have their vertices at the origin (see the previous section), you can extract information about the elements:
  • If 4a is positive, the curve opens right; if 4a is negative, the curve opens left.
  • If  , the parabola is relatively wide; if  , the parabola is relatively narrow.
• Opening up or down: . When the x variable is squared, the parabola opens up or down. Here’s the info you can extract from this equation:
  • If 4a is positive, the parabola opens upward; if 4a is negative, the curve opens downward.
  • If  , the parabola is wide; if  , the parabola is narrow.

The standard forms used for parabolas with their vertices at the origin (which I discuss in the previous sections) are special cases of these more general parabolas. If you replace the h and k coordinates with zeros, you have the special parabolas anchored at the origin.

A move in the position of the vertex (away from the origin, for example) changes the focus, directrix, and axis of symmetry of a parabola. In general, a move of the vertex just adds the value of h or k to the basic form. For instance, when the vertex is at (h, k), the focus is at for parabolas opening to the right and at for parabolas opening upward. The directrix is also affected by h or k in its equation; it becomes for parabolas opening sideways and for those opening up or down. The whole graph shifts its position, but the shift doesn’t affect which direction it opens or how wide it opens. The shape and direction stay the same.

Sketching the graphs of parabolas

Parabolas have distinctive U-shaped graphs, and with just a little information, you can make a relatively accurate sketch of the graph of a particular parabola. The first step is to think of all parabolas as being in one of the general forms I list in the previous two sections. (Refer to the rules in the previous two sections when you graph a parabola.)

Taking the necessary graphing steps

Here’s the full list of steps to follow when sketching the graph of a parabola — either or :

1. Determine the coordinates of the vertex, (h, k), and plot it.

   If the equation contains or , change the forms to or , respectively, to determine the correct signs. Actually, you’re just reversing the sign that’s already there.

2. Determine the direction the parabola opens, and decide if it’s wide or narrow, by looking at the 4a portion of the general parabola equation.
3. Lightly sketch in the axis of symmetry that goes through the vertex ( when the parabola opens upward or downward and when it opens sideways).
4. Choose a couple other points on the parabola and find each of their partners on the other side of the axis of symmetry to help you with the sketch.

For example, if you want to graph the parabola , you first note that this parabola has its vertex at the point and opens to the right, because the y is squared (if the x had been squared, it would open up or down) and a (2) is positive. The graph is relatively wide about the axis of symmetry, , because , which makes greater than 1. To find a random point on the parabola, try letting and solve for becomes . Square the 8 and then divide each side of the equation by 8 to get . Adding 1 to each side, ; so the point on the parabola is (9, 6).

Another example point, which you find by using the same process that gave you (9, 6), is (5.5, 4). Figure 11-5 shows the vertex, axis of symmetry, and the two points placed in a sketch.

The two randomly chosen points have counterparts on the opposite side of the axis of symmetry. The point (9, 6) is 8 units above the axis of symmetry, so 8 units below the axis puts you at . The point (5.5, 4) is 6 units above the axis of symmetry, so its partner is the point .

Applying suspense to the parabola

Sketching parabolas helps you visualize how they’re used in an application, so you want to be able to sketch quickly and accurately if need be. A rather natural occurrence of a curve close to the parabola involves the cables that hang between the towers of a suspension bridge. These cables form a curve called a catenary, which is usually very close to a parabolic curve. Consider the following situation: An electrician wants to put a decorative light on the cable of a suspension bridge at a point 100 feet (horizontally) from where the cable touches the roadway at the middle of the bridge. You can see the electrician’s blueprint in Figure 11-6.

The electrician needs to know how high the cable is at a point 100 feet from the center of the bridge so he can plan his lighting experiment. The towers holding the cable are 80 feet high, and the total length of the bridge is 400 feet.

You can help the electrician solve this problem by writing the equation of the parabola that fits all these parameters. The easiest way to handle the problem is to let the bridge roadbed be the x-axis and the center of the bridge be the origin (0, 0). The origin, therefore, is the vertex of the parabola. The parabola opens upward, so you use the equation of a parabola that opens upward with its vertex at the origin, which is (see the section “Looking at parabolas with vertices at the origin”).

To solve for a, you put the coordinates of the point (200, 80) in the equation. Where do you get these seemingly random numbers? Half the total of 400 feet of bridge is 200 feet. You move 200 feet to the right of the middle of the bridge and 80 feet up to get to the top of the right tower. Replacing the x in the equation with 200 and the y with 80, you get . Dividing each side of the equation by 80, you find that 4a is 500, so the equation of the parabola that represents the cable is .

So, how high is the cable at a point 100 feet from the center? In Figure 11-6, the point 100 feet from the center is to the left, so represents x. A parabola is symmetric about its vertex, so it doesn’t matter whether you use positive or negative 100 to solve this problem. But, sticking to the figure, let in the equation; you get , which becomes . Dividing each side by 500, you get . The cable is 20 feet high at the point where the electrician wants to put the light. He needs a ladder!

Converting parabolic equations to the standard form

When the equation of a parabola appears in standard form, you have all the information you need to graph it or to determine some of its characteristics, such as direction or size. Not all equations come packaged that way, though. You may have to do some work on the equation first to be able to identify anything about the parabola.

The standard form of a parabola is or , where (h, k) is the vertex.

The methods used here to rewrite the equation of a parabola into its standard form also apply when rewriting equations of circles, ellipses, and hyperbolas. (See the last section in this chapter, “Identifying Conics from Their Equations, Standard or Not,” for a more generalized view of changing the forms of conic equations.) The standard forms for conic sections are factored forms that allow you to immediately identify needed information. Different algebra situations call for different standard forms — the form just depends on what you need from the equation.

For instance, if you want to convert the equation into the standard form, you perform the following steps, which contain a method called completing the square (a method you use to solve quadratic equations; refer to Chapter 3 for a review of completing the square):

1. Rewrite the equation with the and x terms (or the and y terms) on one side of the equation and the rest of the terms on the other side.
   

2. Add a number to each side to make the side with the squared term into a perfect square trinomial (thus completing the square; see Chapter 3 for more on trinomials).

   In this case, you add 25 to each side. simplifies to .

3. Rewrite the perfect square trinomial in factored form, and factor the terms on the other side by the coefficient of the variable.
   

You now have the equation in standard form. The vertex is at ; it opens upward and is fairly wide (see the section “Sketching the graphs of parabolas” to see how to make these determinations).

Standardizing the circle

When the equation of a circle appears in the standard form, it provides you with all you need to know about the circle: its center and radius. With these two bits of information, you can sketch the graph of the circle. The equation , for example, is the equation of a circle. You can change this equation to the standard form by completing the square for each of the variables (refer to Chapter 3 if you need a review of this process). Just follow these steps:

1. Change the order of the terms so that the x’s and y’s are grouped together and the constant appears on the other side of the equal sign.

   Leave a space after the groupings for the numbers that you need to add:

   

2. Complete the square for each variable, adding the number that creates perfect square trinomials.

   In the case of the x’s, you add 9, and with the y’s, you add 4. Don’t forget to also add 9 and 4 to the right:

   

   When it’s simplified, you have .

3. Factor each perfect square trinomial.

   The standard form for the equation of this circle is .

The circle has its center at the point and has a radius of 4 (the square root of 16). To sketch this circle, you locate the point and then count 4 units up, down, left, and right; sketch in a circle that includes those points. Figure 11-8 shows you the way.

Specializing in circles

Two circles you should consider special are: the circle with its center at the origin, and the unit circle.

A circle with its center at the origin has, of course, a center at (0, 0), so its standard equation becomes .

The equation of the center origin is simple and easy to work with, so you should take advantage of its simplicity and try to manipulate any application you’re working with that uses a circle into one with its center at the origin.

The unit circle also has its center at the origin, but it always has a radius of one. The equation of the unit circle is . This circle is also convenient and nice to work with. You use it to define trigonometric functions, and you find it in analytic geometry and calculus applications.

For example, a circle with its center at the origin and a radius of 5 units is created from the equation , where (h, k) is (0, 0) and ; therefore, it has the equation . Any circle has an infinite number of points on it, but this clever choice for the radius gives you plenty of integers for coordinates. Points lying on the circle include , and . Not all circles offer this many integral coordinates, which is why this one is one of the favorites. (The rest of the infinite number of points on the circle have coordinates that involve fractions and radicals.)

Raising the standards of an ellipse

You can think of the ellipse as a sort of squished circle. Of course, there’s much more to ellipses than that, but the label sticks because the standard equation of an ellipse has a vague resemblance to the equation for a circle (see the previous section).

The standard equation for an ellipse with its center at the point (h, k) is , where

• x and y are points on the ellipse.
• a is half the length of the ellipse from left to right at its widest point.
• b is half the distance up or down the ellipse at its tallest point.

To be successful in elliptical problems, you need to find out more from the standard equation than just the center. You want to know if the ellipse is long and narrow or tall and slim. How long is it across, and how far up and down does it go? You may even want to know the coordinates of the foci. You can determine all these elements from the equation.

Determining the shape

An ellipse is criss-crossed by a major axis and a minor axis. Each axis divides the ellipse into two equal halves, with the major axis being the longer of the segments (such as the x-axis in Figure 11-9; if no axis is longer, you have a circle). The two axes intersect at the center of the ellipse. At the ends of the major axis, you find the vertices of the ellipse. Figure 11-10 shows two ellipses with their axes and vertices identified.

To determine the shape of an ellipse, you need to pinpoint two characteristics:

• Lengths of the axes: You can determine the lengths of the two axes from the standard equation of the ellipse. You take the square roots of the numbers in the denominators of the fractions. Whichever value is larger, a or b, tells you which is the major axis. The square roots of the numbers in the denominator represent the distances from the center to the points on the ellipse at the end of their respective axes. In other words, a is half the length of one axis, and b is half the length of the other. Therefore, 2a and 2b are the lengths of the axes.

  To find the lengths of the major and minor axes of the ellipse , for example, you take the square roots of 25 and 49. The square root of the larger number, the 49, is 7. Twice 7 is 14, so the major axis is 14 units long. The square root of 25 is 5, and twice 5 is 10. The minor axis is 10 units long.

• Assignment of the axes: The positioning of the axes is significant. The denominator that falls under the x signifies the axis that runs parallel to the x-axis. In the previous example, the 25 falls under the x, so the minor axis runs horizontal. The denominator that falls under the y factor is the axis that runs parallel to the y-axis. In the previous example, 49 falls under the y, so the major axis runs up and down parallel to the y-axis. This is a tall, narrow ellipse.

Finding the foci

You can find the two foci of an ellipse by using information from the standard equation. The foci, for starters, always lie on the major axis. They lie c units from the center. To find the value of c, you use parts of the ellipse equation to form the equation or , depending on which is larger, or . The value of has to be positive.

In the ellipse , for example, the major axis runs across the ellipse, parallel to the x-axis (see the previous section to find out why). Actually, the major axis is on the x-axis, because the center of this ellipse is the origin. You know this because the h and k are missing from the equation (actually, they’re both equal to zero). You find the foci of this ellipse by solving the foci equation: . Substituting 25 for and 9 for , you have . Taking the square root, .

So, the foci are 4 units on either side of the center of the ellipse. In this case, the coordinates of the foci are and (4, 0). Figure 11-11 shows the graph of the ellipse with the foci identified.

Also, for this example ellipse, the major axis is 10 units long, running from to (5, 0). These two points are the vertices. The minor axis is 6 units long, running from (0, 3) down to .

When the center of the ellipse isn’t at the origin, you find the foci the same way and adjust for the center. The ellipse has its center at . You find the foci by solving , which, in this case, is . The value of c is either 24 (the root of 576) or from the center, so the foci are (23, 3) and . The major axis is units, and the minor axis is units. The 25 and 7 come from the square roots of 625 and 49, respectively. And the vertices, the endpoints of the major axis, are at (24, 3) and .

Sketching an elliptical path

Have you ever been in a whispering gallery? I’m talking about a room or auditorium where you can stand at a spot and whisper a message, and a person standing a great distance away from you can hear your message. This phenomenon was much more impressive before the days of hidden microphones, which tend to make us more skeptical of how this works. Anyway, here’s the algebraic principle behind a whispering gallery. You’re standing at one focus of an ellipse, and the other person is standing at the other focus. The sound waves from one focus reflect off the surface or ceiling of the gallery and move over to the other focus.

Suppose you run across a problem on a test that asks you to sketch the ellipse associated with a whispering gallery that has foci 240 feet apart and a major axis (length of the room) of 260 feet. Your first task is to construct the equation of the ellipse.

The foci are 240 feet apart, so they each stand 120 feet from the center of the ellipse. The major axis is 260 feet long, so the vertices are each 130 feet from the center. Using the equation — which gives the relationship between c, the distance of a focus from the center; a, the distance from the center to the end of the major axis; and b, the distance from the center to the end of the minor axis — you get . Armed with the values of and , you can write the equation of the ellipse representing the curvature of the ceiling: .

To sketch the graph of this ellipse, you first locate the center at (0, 0). You count 130 units to the right and left of the center and mark the vertices, and then you count 50 units up and down from the center for the endpoints of the minor axis. You can sketch the ellipse by using these endpoints. The sketch in Figure 11-12 shows the points described and the ellipse. It also shows the foci — where the two people would stand in the whispering gallery.

Including the asymptotes

A very helpful tool you can use to sketch hyperbolas is to first lightly sketch in the two diagonal asymptotes of the hyperbola. Asymptotes aren’t actual parts of the graph; they just help you determine the shape and direction of the curves. The asymptotes of a hyperbola intersect at the center of the hyperbola. You find the equations of the asymptotes by replacing the 1 in the equation of the hyperbola with a zero and simplifying the resulting equation into the equations of two lines.

If you want to find the equations of the asymptotes of the hyperbola , for example, you change the one to zero and then set the two fractions equal to one another:

is then written .

Take the square root of each side:

simplifies to .

Now you multiply each side by 4 to get the equations of the asymptotes in better form:

becomes .

Consider the two cases — one using the positive sign, and the other using the negative sign. When using the positive sign, giving you or .

And, with the negative sign, becomes . Adding 4 to each side, you get or .

The two asymptotes you find are and , and you can see them in Figure 11-14. Notice that the slopes of the lines are the opposites of one another. (For a refresher on graphing lines, see Chapter 2.)

Graphing hyperbolas

Hyperbolas are relatively easy to sketch, if you pick up the necessary information from the equations (see the intro to this section). To graph a hyperbola, use the following steps as guidelines:

1. Determine if the hyperbola opens to the sides or up and down by noting whether the x term is first or second.

   The x term first means it opens to the sides.

2. Find the center of the hyperbola by looking at the values of h and k.

3. Lightly sketch in a rectangle twice as wide as the square root of the denominator under the x value and twice as high as the square root of the denominator under the y value.

   The rectangle’s center is the center of the hyperbola.

4. Lightly sketch in the asymptotes through the vertices of the rectangle (see the previous section to find out how).
5. Draw in the hyperbola, making sure it touches the midpoints of the sides of the rectangle.

You can use these steps to graph the hyperbola . First, note that this equation opens to the left and right because the x value comes first in the equation. The center of the hyperbola is at .

Now comes the mysterious rectangle. In Figure 11-15a, you see the center placed on the graph at . You count 3 units to the right and left of center (totaling 6), because twice the square root of 9 is 6. Now you count 4 units up and down from center, because twice the square root of 16 is 8. A rectangle 6 units wide and 8 units high is shown in Figure 11-15b.

When the rectangle is in place, you draw in the asymptotes of the hyperbola diagonally through the vertices (corners) of the rectangle. Figure 11-16a shows the asymptotes drawn in. The equations of those asymptotes are and (see the previous section to calculate these equations). Note: When you’re just sketching the hyperbola, you usually don’t need the equations of the asymptotes.

Lastly, with the asymptotes in place, you draw in the hyperbola, making sure it touches the sides of the rectangle at its midpoints and slowly gets closer and closer to the asymptotes as the curves get farther from the center. You can see the full hyperbola in Figure 11-16b.