Chapter 10 Conic Sections © 2012 McGraw-Hill Companies, Inc. All rights reserved.

CircleEllipseHyperbola In this chapter, we will study the conic sections. When a right circular cone is intersected by a plane, the result is a conic section. The conic sections are parabolas, circles, ellipses, and hyperbolas. The following figures show how each conic section is obtained from the intersection of a cone and a plane. Parabola In Chapter 8, we learned how to graph parabolas. The graph of a quadratic function, f(x)=ax 2 +bx +c, is a parabola that opens vertically. Another form this function may take is f(x)=a(x – h) 2 +k. The graph of a quadratic equation of the form x=ay 2 + by + c, or x = a(y – k ) 2 + h, is a parabola that opens horizontally. The next conic section we will discuss is the circle. Define a Conic Section

The midpoint of a diameter of a circle is the center of the circle. Graph a Circle Given in the Form (x – h ) 2 + (y – k ) 2 = r 2 A circle is defined as the set of all points in a plane equidistant (the same distance) from a fixed point. The fixed point is the center of the circle. The distance from the center to a point on the circle is the radius of the circle. Let the center of a circle have coordinates ( h, k ) and let ( x, y ) represent any point on the circle. Let r represent the distance between these two points, r is the radius of the circle. We will use the distance formula to find the distance between the center, ( h, k), and the point ( x, y ) on the circle. We will use the distance formula to find the distance between the center, ( h, k ), and the point ( x, y ) on the circle. Substitute (x, y) for (x 2, y 2 ), (h, k) for (x 1, y 1 ), and r for d. Distance Formula Square both sides.

This is the standard form for the equation of a circle. Example 2 Graph (x + 1) 2 + (y + 2) 2 = 4. Solution Standard form is (x – h) 2 + (y – k) 2 = r 2. Our equation is (x + 1) 2 + (y + 2) 2 = 4. h = -1 k = -2 r = = 2 The center is (-1, -2). The radius is 2. To graph the circle, first plot the center (-1,-2) use the radius to locate four points on the circle. From the center, move 2 units up, down, left, and right. Draw a circle through the four points.

Example 3 Graph x 2 + y 2 = 1. Solution Standard form is (x – h) 2 + (y – k) 2 = r 2. Our equation is x 2 + y 2 = 1. h = 0 k = 0 r = = 1 The center is (0,0). The radius is 1. To graph the circle, first plot the center (0,0) use the radius to locate four points on the circle. From the center, move 1 unit up, down, left, and right. Draw a circle through the four points. The circle x 2 + y 2 = 1 is used often in other areas of mathematics such as trigonometry. x 2 + y 2 = 1 is called the unit circle.

If we are told the center and radius of a circle, we can write its equation. Example 4 Find an equation of the circle with center (3, -2) and radius. Solution The x-coordinate of the center is h. What is h? The y-coordinate of the center is k. What is k? What is r? h = 3 k = -2 r = Substitute these values into(x – h) 2 + (y – k) 2 = r 2 [x – (3)] 2 + (y – -2) 2 = ( ) 2 (x – 3) 2 + (y + 2) 2 = 6 Substitute 3 for x, -2 for k, and for r. Equation

Graph a Circle of the Form Ax 2 + Ay 2 + Cx + Dy + E = 0 The equation of a circle can take another form—general form. To graph a circle given in this form, we complete the square on x and on y to put it into standard form. After we learn all of the conic sections, it is very important that we understand how to identify each one. To do this, we will usually look at the coefficients of the square terms.

Example 5 Graph The coefficients of x 2 and y 2 are each 1. Therefore, this is the equation of a circle. Our goal is to write the given equation in standard form, (x – h) 2 + (y – k) 2 = r 2, so that we can identify its center and radius. To do this, we will group x 2 and 8x together, group y 2 and 4y together, then complete the square on each group of terms. Solution Group x 2 and 8x together. Group y 2 and 4y together. Move the constant to the other side. Complete the square for each group of terms. (x 2 + 8x + 16) + (y 2 +4y +4) = (x+4) 2 + (y+2) 2 = 7 The center of the circle is (– 4, – 2 ). The radius is Since 16 and 4 are added on the left, they must also be added on the right. Factor.

The center of the circle is (– 4, – 2 ). The radius is (– 4, – 2 ) First plot the center. (– 4, – 2 ) Second use the radius to plot more points on the circle. From the center go units To the left, right, up, and down. y x

The next conic section we will study is the ellipse. An ellipse is the set of all points in a plane such that the sum of the distances from a point on the ellipse to two fixed points is constant. Each fixed point is called a focus (plural: foci). The point halfway between the foci is the center of the ellipse. The Ellipse Graph an Ellipse

The orbits of planets around the sun as well as satellites around the earth are elliptical. Statuary Hall in the U.S. Capitol building is an ellipse. If a person stands at one focus of this ellipse and whispers, a person standing across the room on the other focus can clearly hear what was said. Properties of the ellipse are used in medicine as well. One procedure for treating kidney stones involves immersing the patient in an elliptical tub of water. The kidney stone is at one focus, while at the other focus, high energy shock waves are produced, which destroy the kidney stone.

Graph Example 1 Solution Standard form is Our equation is What is h ?What is k ? What is a ? What is b ? h = -4 a=a= b = Since a = 5 and a 2 is under the squared quantity containing the x, move 5 units each way in the x – direction from the center. These are two points on the ellipse. k = -1 (– 4, – 1 ) y x The center is (-4, -1). To graph the ellipse, first plot the center (-4,-1). Since b = 3 and b 2 is under the squared quantity containing y, move 3 units each way in the y – direction from the center. These are two more points on the ellipse

Graph Example 2 Solution Since a = 2 and a 2 is under the squared quantity containing the x, move 2 units each way in the x – direction from the center. These are two points on the ellipse. Standard form is Our equation is What is h ?What is k ? What is a ?What is b ? h = 0 a=a= b = k =0 The center is (0, 0). To graph the ellipse, first plot the center (0,0). Since b = 4and b 2 is under the squared quantity containing y, move 4 units each way in the y – direction from the center. These are two more points on the ellipse. (0, 0 ) y x (0, – 4 ) (0, 4) (2, 0)(– 2, 0)

In Example 2, note that the origin, (0,0), is the center of the ellipse. Notice also that a = 2 and the x-intercepts are (2,0) and (-2,0); b = 4 and the y-intercepts are (0,4) and (0, -4). We can generalize these relationships as follows.

Looking at Examples 1 and 2, we can make another interesting observation. Example 1 a 2 = 25 b 2 = 9 a 2 > b 2 a 2 = 4 b 2 = 16 b 2 > a 2 The number under ( x +4) 2 is greater than the number under ( y + 1) 2. The ellipse is longer in the x -direction. The number under y 2 is greater than the number under x 2. The ellipse is longer in the y -direction. This relationship between a 2 and b 2 will always produce the same result. The equation of an ellipse can take other forms. Example 2

Graph Example 3 Solution How can we tell whether this is a circle or an ellipse? We look at the coefficients of x 2 and y 2. Both of the coefficients are positive, and they are different. This is an ellipse. (If this were a circle, the coefficients would be the same.) Since the standard form for the equation of an ellipse has a 1 on one side of the sign, divide both sides of by 144 to obtain a 1 on the right. Divide both sides by 144. Simplify. The center is (0, 0). What is a ?What is b ? a=a= b = (0, 0 ) y x (0, – 3 ) (0,3) 4, 0) (– 4, 0)

The Hyperbola Graph a Hyperbola The last of the conic sections is the hyperbola. A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant. Each fixed point is called a focus. The point halfway between the foci is the center of the hyperbola. Some navigation systems used by ships are based on the properties of hyperbolas. A lamp casts a hyperbolic shadow on a wall, and many telescopes use hyperbolic lenses. A hyperbola is a graph consisting of two branches. The hyperbolas we will consider will have branches that open either in the x-direction or in the y-direction.

Notice how the branches of the hyperbola get closer to the dotted lines as the branches continue indefinitely. These dotted lines are called Asymptotes.

Graph Example 4 Solution How do we know that this is a hyperbola and not an ellipse? It is a hyperbola because there is subtraction sign between the two quantities on the left. If it were addition, it would be an ellipse. Standard form is Our equation is What is h ?What is k ? What is a ?What is b ? h = 3 a=a= b = k = – 2 The center is (3, -2).

Use the center, ( 3, -2 ), a = 2, b = 3 to draw a reference rectangle. The diagonals of this rectangle are the asymptotes of the hyperbola. First, plot the center (3, -2). Since a = 2 and a 2 is under the squared quantity containing the x, move 2 units each way in the x- direction from the center. These are two points on the rectangle. Since b = 3 and b 2 is under the squared quantity containing the y, move 3 units each way in the y -direction from the center. These are two more points on the rectangle. Draw the rectangle containing these four points, then draw the diagonals of the rectangle as dotted lines. These are the asymptotes of the hyperbola. Sketch the branches of the hyperbola opening in the x- direction with the branches approaching the asymptotes. diagonals

Graph Example 5 Standard form is Our equation is What is h ?What is k ? What is a? What is b ? h = 0 a=a= b = k = 0 The center is (0, 0). Solution

Graph Example 6 Solution This is a hyperbola since there is a subtraction sign between the two terms. Since the standard form for the equation of the hyperbola has a 1 on one side of the equal sign, divide both side of y 2 – 9x 2 = 9 by 9 to obtain a 1 on the right. The center is (0, 0). Divide both sides by 9. Simplify. h = 0 a=a= b = k = 0