Figures adapted from Mathworld.wolfram.com and vectosite.net.

Transcription

1 MTH 11 CONIC SECTIONS 1 The four basic types of conic sections we will discuss are: circles, parabolas, ellipses, and hyperbolas. They were named conic by the Greeks who used them to describe the intersection of a plane with a double cone. Although you may have seen equations for some of these curves in the past, this time we will be approaching them from a different point of view. Figures adapted from Mathworld.wolfram.com and vectosite.net. I. CIRCLES A circle is the set of points in a plane equidistant from a fixed point, called the center. The distance from any point on the circle to the center is called the radius. If the center is C(h, k), and the radius is r, then the distance from C to any other point, (x, y) on the circle is found by using the distance formula. ( x h) + ( y k) = r If we square both sides of this equation we get ( x h) + ( y k) = r Which is the standard equation of a circle we have used in the past. Example 1: Find the standard equation of a circle with radius and center (4, 3). Then sketch the graph. We are given r =, h = 4, and k = 3, so the equation of the circle is ( x 4) + ( y + 3) = Linn-Benton Community College MTH 11 Conics -6

2 If we multiply the standard form of a circle out and clear out the fractions, we will get an equation of the general quadratic form: Ax + By + Cx + Dy + E = 0. If A = B, this will be a circle. Example : Put the following equation of a circle in standard form. State the center and radius. x + y + 8x 4y = 0 Since there are coefficients of the squared terms that are not 1, we will start by dividing each term in the equation by. x y 8x 4y = x + y + 4x y 11 = 0 Now we rearrange the terms and complete the square on x and on y. x + 4x + y y = 11 x x y y = ( + ) + ( 1) = 16 x y The equation is in standard form. The center is (,1), and the radius is r = 4. Example 3: Write the equation of the circle with endpoints of the diameter at (,1) and (4, 3). The center of the circle must be the midpoint of the diameter. To find it we will put 1 1 the diameter endpoints into the midpoint formula x + x, y + y to get ( 3), =, = (1, 1). This is the center of the circle and will be substituted for ( h, k ) in the standard form for a circle. Let h = 1 and k = 1. ( x h) + ( y k) = r ( x 1) + ( y ( 1)) = r ( x 1) + ( y + 1) = r To get the radius, r, we note that the circle must pass through both of the endpoints on the diameter. We can use one of them and substitute the values for x and y in the equation and solve for r. Use the point (,1). Let x = and y = 1. ( 1) + (1 + 1) = r ( 3) + () = r = r 13 = r 13 = r Now we can write the standard form of the equation of this circle. ( x 1) + ( y + 1) = Linn-Benton Community College MTH 11 Conics

3 EERCISES: 1. Find the equation of the circle with radius 9 and center ( 3, 1).. Find the equation of the circle with radius 7 and center (4, ). 3. a. Draw the set of points which are 5 units from (, 3). b. Find an equation for the set of points. 4. a. Draw the set of points which are 7 units from (0, 0). b. Find an equation for the set of points. For Exercises 5 and 6, identify the center and radius of the given circles. 5. ( x + 8) + ( y 11) = 1 6. ( x 6) + ( y 5) = 18 For Exercises 7 and 8, identify the center and radius, and sketch the graph of the circle. 7. ( x + ) + ( y 3) = ( x 6) + y = 36 For Exercises 9 through 1, find the center and radius, and then sketch the graph of each circle. 9. x + y + 8y + 14 = = 0 x y y = 0 x y x y = 0 x y x y For Exercises 13 through 16, find the equation of a circle that satisfies the given information. 13. The diameter of the circle passes through the points ( 5, 6) and (7, 4). 14. The diameter of the circle passes through the points ( 3, ) and (3, 4). 15. The center of a circle is at (7, 5) and a point on the circle is (, 3). 16. The center of a circle is at (0, 8) and a point on the circle is (9, 0) Linn-Benton Community College MTH 11 Conics

4 II. PARABOLAS 4 A parabola is defined as the set of all points P(x, y) in a plane which are equidistant from a fixed line called the directrix and a fixed point (not on the line) called the focus. If the vertex of a parabola opening up or down is at (0,0), the focus is at (0, p), then the directrix is y = p. P(x, y) focus (0, p) directrix y = p If P(x, y) is any point on the parabola. Then, by the definition of a parabola given above, P(x, y) is equidistant from both the focus (0, p) and the directrix, y = p. The point on the directrix closest to (x, y) is (x, p). Let d 1 be the distance from P(x, y) to (0, p) and let d be the distance from P(x, y) to (x, p) as shown in the next figure. d 1 P(x, y) focus (0, p) d (x, p) By the definition of a parabola, d 1 = d. Using the distance formula, we have ( x 0) + ( y p) = ( x x) + ( y ( p)) or x + y py + p = 0 + y + py + p Simplify and square both sides x + y py + p = y + py + p Now subtracting y, p, and adding py to both sides leaves x = 4 py. directrix y = p This is the standard form of a parabola that opens up or down and has its vertex at (0, 0), focus at (0, p), and directrix, y = p. Recall that the axis of symmetry for a parabola passes through the vertex. Notice that the focus is on the axis of symmetry for the parabola and the directrix is perpendicular to the axis of symmetry. 007 Linn-Benton Community College MTH 11 Conics

5 Standard Form of a Parabola: Vertex at (0, 0), focus at (0, p), and directrix at y = p. 5 x = 4 py If p > 0, the parabola opens up. If p < 0, the parabola opens down. Example 1: State the coordinates of the vertex and the focus, and the equation of the directrix of the parabola x = y. Then sketch the graph. First find p. Compare the standard form of a parabola to the given equation and note that 4p =, so p = 1 = 0.5. The vertex is (0, 0), so the focus is p = 0.5 units up along the axis of symmetry at (0, 0.5), and the directrix is y = 0.5. Because p is positive, this parabola opens up. To determine how wide to sketch the parabola, select a y-value and use the equation to find the x-values that are paired with it. Select y =, for example, and substitute y = into x = y, then solve for x. x x = () = 4 x = ± 4 x = ± Plot the points (, ) and (, ) and complete the sketch of the parabola. 4 focus (0, 0.5) - 0 y = Example : State the coordinates of the vertex and the focus, and the equation of the directrix of the parabola x = 1y. Then sketch the graph. Find p: 4p = 1, so p = 3. Because p is negative, this parabola opens down. The vertex is (0, 0), focus is (0, 3), and directrix is y = 3. Plot two additional points by selecting a y-value, say y = 3, substitute it into the equation of the parabola and solve for x. x = 1 3 = 36, and x = ± 6. Plot (6, 3) and ( 6, 3). If y = 3, then ( ) 007 Linn-Benton Community College MTH 11 Conics

6 6 4 3 y = focus (0, 3) -4 We have seen that a parabola with general form, x = 4 py, opens up if p > 0 and opens down if p < 0. A parabola can also open left or right. A parabola with vertex at (0, 0) which opens left or right has the general form: y = 4 px. If p > 0, the parabola opens right as shown. If p < 0, the parabola opens left as shown. focus (p, 0) focus ( p, 0) directrix x = p directrix x = p Standard Form of a Parabola: Vertex at (0, 0), focus at (p, 0), and directrix at y = p. y = 4 px If p > 0, the parabola opens right. If p < 0, the parabola opens left. 007 Linn-Benton Community College MTH 11 Conics

7 7 The standard form of a parabola with vertex at (h, k) can be found by translating the forms we have seen already from vertex (0, 0) to vertex ( h, k ) by substituting x h for x and y k for y to get: Opens up or down: (x h) = 4p(y k) with focus F(h + p, k) and directrix y = p. Opens right or left: (y k) = 4p(x h) with focus F(h, k + p) and directrix x = p. Example 3: State the coordinates of the vertex and focus, and the equation of the directrix, then sketch the graph of the parabola (x + ) = 8(y 3). 4p = 8 so p =. The vertex is at V(, 3) and the focus is at F(, 3 + ) or F(, 5). The directrix is the line y = 3, or y = 1. Plot two additional points. If y = 5, then ( x + ) = 8(5 3) + = ( x ) 16 x + = ± 4 x = ± 4 x = 6 or Plot the points (, 5) and ( 6, 5). F V y = 1 In Example 3, we used the y-coordinate of the focus to generate two points on the parabola. Those points lie to the right and to the left the focus. The line segment joining those two points that passes through the focus is called the latus rectum. Example 4: State the coordinates of the vertex and focus, and the equation of the directrix, then sketch the graph of the parabola (y 1) = 4(x ). Find the endpoints of the latus rectum. 4p = 4 so p = 1. Because p < 0, this parabola opens left. The vertex is V(, 1). The focus is F( 1, 1) or F(1, 1). The directrix is the line x = ( 1), or x =3. When x = 1, ( y 1) = 4(1 ), so y = 3 or 1. Plot (1, 3) and (1, 1). These are the endpoints of the latus rectum. The latus rectum is shown as a dashed segment in the figure. 4 F V - x = Linn-Benton Community College MTH 11 Conics

8 Parabolas with axes of symmetry parallel to a coordinate axis and vertex (h, k) p Focus Directrix Equation Description p > 0 (h + p, k) x = h p (y k) = 4p(x h) Axis of symmetry parallel to x-axis, opens right p < 0 (h + p, k) x = h p (y k) = 4p(x h) Axis of symmetry parallel to x-axis, opens left p > 0 (h, k + p) y = k p (x h) = 4p(y k) Axis of symmetry parallel to y-axis, opens up p < 0 (h, k + p) y = k p (x h) = 4p(y k) Axis of symmetry parallel to y-axis, opens down 8 If the equation of a parabola is put into the general quadratic form with integer coefficients it will be of the form Ax + Bx + Cy + D = 0 or Ay + Bx + Cy + D = 0. Notice that the equations do not have both x and y terms. Using completing the square the equations can be put into the standard form for a parabola. EERCISES: For Exercises 1 through 8 state the coordinates of the vertex and focus, the equation of the directrix, and sketch the graph. Also find and label the endpoints of the latus rectum. 1. y = 4x. y = 6x 3. x = 8y 4. x = 16y 5. (y 1) = 0(x + ) 6. (y 3) = 1(x + 1) 7. (x 4) = 4(y + 3) 8. (x 5) = 4(y ) 9. Sketch all the points that are equidistant from the line x = and the point (, 3). Find an equation for this curve. 10. Sketch all the points that are equidistant from the line y = 6 and the point ( 1, 1). Find an equation for this curve. 11. Find the equation of the parabola with the focus at (, 5) and directrix y = Find the equation of the parabola with focus at ( 4, ) and directrix x = 3. Exercises 13 and 14: The surface formed by rotating a parabola about its axis of symmetry is called a paraboloid of revolution. If the paraboloid of revolution is sliced through its axis of symmetry, the cross section is a parabola. 13. The reflective surface of a good flashlight is in the shape of a paraboloid of revolution. Any light radiating from a light bulb set at the focus will reflect off the mirrored surface as parallel beams of light. (See the figure.) If the light bulb is set 1. cm from the base of the surface, find the equation of the parabolic cross section. Hint: Place the vertex at (0, 0). How wide is the flashlight at its widest point if it is 4 cm deep at the center? Round to the nearest tenth of a centimeter. Focus 14. A satellite dish is in the shape of a paraboloid of revolution. Signals that hit the surface of the satellite dish are reflected to one point, the focus, where the receiver is located. At its widest point, the satellite dish is 10 feet wide. It is 3.5 feet deep at its center. Determine the where the receiver should be placed. Hint: Find 007 Linn-Benton Community College MTH 11 Conics

9 the equation of the parabolic cross section and assume the base of the satellite dish is at (0, 0). 9 For Problems 17 and 18, identify whether the equation represents a circle, parabola or neither. If it is a circle or parabola, put it into standard form. 17. a. x + y + 4x 6y 3 = 0 b. c. x x y = = x y 4x An arched doorway in the shape of a parabola is 1 feet at its tallest point. It is 4 feet wide at the base. How wide is the doorway at a height of 6 feet above the floor? Round to the nearest tenth of a foot. Hint: Find an equation of the parabola first. 18. a. b. c. x y x y = 0 y x y = 0 x + y + 1x 6 = A bridge built in 1914 was constructed with a parabolic hinged-rib arch. The arch is 58 feet across at its widest point, where it intersects the concrete walls at a point 10 feet above the sidewalk. The height from the pavement to the highest part of the arch is 35 feet. How wide is the arch 15 feet above the pavement? Round to the nearest tenth of a foot. Hint: Find an equation of the parabola first. 007 Linn-Benton Community College MTH 11 Conics

10 III. ELLIPSES 10 An ellipse is the set of points in a plane, the sum of whose distances from two fixed points, called the foci, is a constant. If we let the sum of the two distances, F 1 P + F P = a where a is a constant, then the ellipse with center at the origin, longest axis (called the major axis) along the x- axis, shorter axis (minor axis) along the y-axis has the following standard equation. x a y + = where b 1 (0, b) a b = c Note: The relationship a b = c does not match the Pythagorean theorem. And that in this equation a > b. The graph of an ellipse that matches this equation will have the following general form. The points at the ends of the major axes are called vertices, in the graph these are labeled V 1 and V. V 1 ( a, 0) F 1 ( c, 0) F (c, 0) V (a, 0) (0, b) If the center is at the origin, and the two foci are on the y-axis, the major axis will be vertical and the minor axis will be horizontal. Then the standard equation will be y x + = 1 where a b = c a b And a graph might look like V 1 (0, a) F 1 (0, c) ( b, 0) (b, 0) F (0, c) V (0, a) 007 Linn-Benton Community College MTH 11 Conics

11 11 Note: ou can tell whether the major axis of an ellipse is horizontal or vertical by comparing the values underneath the x and y terms. If the largest value is below x, the major axis is horizontal. If the largest value is below y, the major axis is vertical. Example 1: Find the coordinates of the vertices and foci and sketch: 5x + 3y = 15. x y y x We have to put this into the standard form + = 1 or + = 1 first. Create a a b a b 1 on the right-hand side of the equation by dividing each term in the equation by 15. 5x 3y 15 x y + =, which simplifies to + = Since 5 > 3, and 5 is the denominator of the y term, the major axis is vertical. a = 5, so a = 5 ; and b = 3, so b = 3. Since a b = c, c = 5 3 and c =. The center is C(0, 0). The vertices and foci are equidistant from the center. The vertices occur at V 1 ( 0, 5 ) and V ( 0, 5 ). The foci occur at F 1 ( 0, ) and F ( 0, ). Also note that the endpoints of the minor axis are at ( 3,0 ) and ( 3,0), however, these points are not called vertices. They are used for graphing purposes only Example : Find the equation of an ellipse, centered at the origin with V 1 ( 6,0) and one focus at (, 0). Also give the coordinates of the other vertex and focus. - V 1 F F V -3 Notice that the given vertex and focus points are both on the x-axis, so the major axis is horizontal. Thus, a must be the denominator of the x term in the equation. The value of a is defined as the distance from the center to a vertex point. The distance from the coordinates of V 1 to the origin is 6 units, so a = 6. The value of c is defined as the distance from the center to a focus point. The distance from (, 0) to the origin is units, so c =. Since a b = c, we have 36 b = 4, so b = 36 4 = 3. The x y equation is + = 1. Since the vertex points and the focus points are equidistant 36 3 from the center, the other vertex is at V (6, 0) and the other focus is at (, 0). 007 Linn-Benton Community College MTH 11 Conics

12 1 To translate the center of an ellipse from (0, 0) to any point ( h, k) we substitute x h for x and y k for y to get: Standard Form of an Ellipse Center (h, k), a > b ( x h) ( y k) + = 1 Major Axis is Horizontal a b ( y k) ( x h) + = 1 Major Axis is Vertical a b Example 3: Find the coordinates of the center, vertices and foci for sketch the graph. ( x ) ( y + 3) + = 1, and 9 4 The center is C(, 3). The major axis is horizontal since the larger number is in the denominator of the x term. Thus, a = 9 and a = 3; b = 4 and b =. Find the value for c by using the values of a and b: c = a b = 9 4 = 5, so c = 5. The vertices are found by adding or subtracting the value of a from the x-coordinate of the center. Thus, the vertices are ( ± 3, 3), or V 1 ( 1, 3) and V (5, 3). Similarly, the foci are ( ± 5, 3), or F 1 ( 5, 3) and F ( + 5, 3). To graph the ellipse, locate the endpoints of the minor axis at (, 3 ± ), or (, 5) and (, 1) V 1 F 1 C(, 3) F V Linn-Benton Community College MTH 11 Conics

13 13 Example 4: Find the equation of an ellipse with center at ( 4,) and major axis parallel to the y- axis, length of major axis equal to 8, and length of minor axis equal to 6. Since the length of the major axis is a, a = 4. The length of the minor axis is b, so b = 3. The desired ellipse has the following equation. ( y ) ( x + 4) + = An ellipse has a very special reflective property. If a source of sound is placed at one focus of the ellipse, then the sound wave will reflect off the ellipse and travel to the other focus. The interior of the dome of St. Paul s Cathedral in London is built with elliptical ceilings, so a whisper uttered at one focus can be clearly heard at the other. A room built in such a way is called a Whispering Gallery. If the ellipse were made up of a reflective surface, then light rays emitted from one focus would reflect to the other focus. In medicine, a machine constructed in the shape of a half ellipsoid produces shock waves from one focus that can break up a kidney stone at the other focus, without surgery. Example 5: A pool table will be constructed in the shape of an ellipse, 50 inches wide and 100 inches long. How far from the center should the cue ball and pocket be placed, so that when the cue ball is hit, it will carom off the wall and land in the pocket? The major axis of the ellipse is 100 inches long and the minor axis is 50 inches long, thus a = 50 and b = 5. If we imagine placing the center of the ellipse at the origin of a coordinate system, with the major axis horizontal, then we can find the locations of the foci, which are where the cue ball and pocket must be placed to complete the desired shot. Find c: c = a b = 50 5 = 1875, thus c = inches. Place the cue ball and pocket along the major axis 43.3 inches to the right and left of the center. 30 cue ball 0 10 pocket Linn-Benton Community College MTH 11 Conics

14 Ellipse with center at (h, k) and major axis horizontal or vertical Center Major axis Foci Vertices Equation (h, k) Horizontal (h + c, k) (h + a, k) ( x h) ( y k) + = 1 a b (h c, k) (h a, k) a > b, a b = c (h, k) Vertical (h, k + c) (h, k + a) ( y k) ( x h) + = 1 a b (h, k c) (h, k a) a > b, a b = c 14 So far we have seen the equation of an ellipse given in standard form. But any equation of the form Ax + By + Cx + Dy + E = 0 in which A B and A and B have the same sign will be an ellipse. Example 6: Find the coordinates of the center and orientation of the major axis for the ellipse given by 5x + 4y + 50x 4y 339 = 0. We need to complete the square on the variables x and y to put the equation in standard form. To set the equation up, we will factor the coefficient of x from the x- terms and the coefficient of y from the y-terms, then complete the square. 5x + 4y + 50x 4y 339 = 0 5( x + x ) + 4( y 6 y ) = 339 x x y y 5( + + 1) + 4( 6 + 9) = x 5( + 1) + 4( 3) = 400 x y + = ( x + 1) ( y 3) + = ( + 1) 4( 3) 400 y The center of the ellipse is the point C( 1,3). And since the larger denominator is under the y variable the major axis is vertical. EERCISES: For Exercises 1 through 6, find the coordinates of the center, vertices, and foci, and sketch the graph. x y 1. + = x y + = x + 5y = x + 6y = 60 ( x 1) ( y + ) + = ( x 3) ( y + 1) + = Linn-Benton Community College MTH 11 Conics

15 For Exercises 7 through 14, find the equation of the ellipse satisfying the given information. 7. Center at (0, 0), focus at ( 1, 0), vertex at (3, 0) 8. Center at (0, 0), focus at (0, 3), vertex at (0, 7) 9. Center at (4, 5), length of horizontal major axis is 10, length of minor axis is Center at (, 3), one focus at (, 0), length of minor axis is Foci at (0, ± 3), x-intercepts are (±, 0) 1. Foci are at (, 0) and (8, 0), length of minor axis is 6 15 Exercises 17 through 0: The orbits of the planets in our solar system are elliptical with the sun at one focus. Some orbits are nearly circular, while others are more elongated, or eccentric. Eccentricity, e, of an ellipse is defined as c. For a circle, eccentricity is a zero. Eccentricities approach 1 as ellipses grow longer and narrower. 17. At perihelion, the Earth is closest to the sun, a distance of million miles. The Earth is farthest away from the sun at aphelion, a distance of 15.6 million miles. Both distances are measured along the major axis of the ellipse as shown. Find the eccentricity of the Earth s orbit. Round to the nearest thousandth. 13. Foci at (±, 0), sum of the distances from the foci to any point is Foci at (0, ±5), sum of the distances from the foci to any point is 1 aphelion perihelion 15. Determine if the quadratic equation given represents an ellipse. If it does, then change it to standard form, state its center, and whether the major axis is horizontal or vertical. a. 4x + 9y 36 = 0 b. c. x y x y = 0 4x 8x + 15y 5 = Determine if the quadratic equation given represents an ellipse. If it does, then change it to standard form, state its center, and whether the major axis is horizontal or vertical. a. 3x + y 1x + 6y 15 = 0 b. c. 6x 5y 150 = 0 9x + 64y 90x + 18y 87 = The length of the major axis of Neptune s orbit is about,65,363,900 km and the length of the minor axis is about,65,1,800 km. Find the eccentricity of Neptune s orbit. Round to the nearest thousandth. 19. Mercury s orbit is more eccentric, e = 0.056, than that of the Earth. At aphelion, Mercury is 69.8 million miles away from the sun. Find Mercury s perihelion distance. Round to the nearest million miles. See Exercise 17 for definitions. 0. The length of Mars minor axis is 7.9 million km. The eccentricity of the orbit of Mars is Find the length of the major axis of Mars orbit. Round to the nearest million kilometers. 007 Linn-Benton Community College MTH 11 Conics

16 1. A brick wall has an opening in the shape of an ellipse, as shown in the following figure. At its widest points, the opening is 4 feet tall and.5 feet wide. ou would like to pass a rectangular box through the window that measures 3.5 feet tall, 1.5 feet wide, and 5 feet deep. The opening in the wall is tall enough to accommodate the 3.5 foot height, but is it wide enough to allow the box through? Explain. Hint: First find the equation of the ellipse by placing the center of the ellipse at the origin of a coordinate system. 16. A bridge is built so that the arch is in the shape of a semi-ellipse. See the following figures. If the major axis of the ellipse is 35 feet wide and the minor axis is 14 feet tall, then how wide is the ellipse at a point 3 feet from the highest point in the arch? Hint: First find the equation of the ellipse by placing the center of the ellipse at the origin of a coordinate system. Round to the nearest tenth. Semi-elliptical Arched Bridge Linn-Benton Community College MTH 11 Conics

17 IV. Hyperbolas 17 A hyperbola is the set of points in a plane such that the difference of the distances from two fixed points, called foci, is a positive constant. The line that joins the two foci, F 1 and F, is called the transverse axis. The vertices, V 1 and V, are located where the hyperbola intersects the transverse axis. The center of the hyperbola is located along the transverse axis halfway between the foci. A possible graph of a hyperbola with a horizontal transverse axis is shown. In the figure P(x, y) represents any point on the hyperbola. P(x, y) F 1 ( c, 0) V1 ( a, 0) V (a, 0) F (c, 0) Transverse Axis Center To create an equation for the hyperbola, the difference in the distances from F 1 to P and F to P must be some positive constant, but what is the constant? Notice that if we let P be one of the vertex points, then the difference in the distances would exactly equal a. With a as the constant, we will write down the difference in the distances, using the distance formula. The distance from F 1 to P minus the distance from F to P is a. x c y x c y a ( ( )) + ( 0) ( ) + ( 0) = ( x + c) + y = a + ( x c) + y ( ( x + c) + y ) = ( a + ( x c) + y ) ( x + c) + y = 4a + 4 a ( x c) + y + ( x c) + y x + cx + c + y = 4a + 4 a ( x c) + y + x cx + c + y 4cx 4a = 4 a ( x c) + y cx a = a ( x c) + y ( ) ( cx a ) = a ( x c) + y 4 c x a cx a a x cx ( + y ) + = + c Isolate one radical expression Square both sides Square the binomials Combine like terms Divide each term by 4 Square both sides Square the binomials 007 Linn-Benton Community College MTH 11 Conics

18 c x a cx + a = a x a cx + a c + a y 4 c a x a y a c a ( ) = ( ) Distribute Collect like terms and factor 18 To simplify the equation further, notice the quantity c a occurs in two places. Refer to the graph of the hyperbola and verify that c is the distance from the center to a focus and a is the distance from the center to a vertex. From that relationship we note that c > a, so c a must be a positive number, let's call it b, and in the equation make the substitution b = c a. = ( c a ) x a y a ( c a ) b x a y = a b b x a y a b = a b a b a b x y = 1 a b Substitute b = c a Divide each term by a b And this final equation is the standard form of the equation of a hyperbola with its center at (0, 0) and a horizontal transverse axis. Look back at the graph of the hyperbola and note that if b = c a, then c = a + b, and there is a geometric meaning for b. Every hyperbola has a fundamental rectangle, which you can see sketched in the following figure. (0, b) F 1 ( c, 0) V1 ( a, 0) V (a, 0) F (c, 0) (0, b) The rectangle is such that its extended diagonals are the asymptotes of the hyperbola. The rectangle has length a and width b. When graphing a hyperbola, be sure to sketch the fundamental rectangle and its diagonals and use them for asymptotes for the hyperbola. The hyperbola we have looked at so far has its center at the origin, (0, 0). To translate the center to any other point (h, k) in the xy-plane we substitute x h for x and y k for y to get the following. 007 Linn-Benton Community College MTH 11 Conics

19 The Standard Form of a Hyperbola with a Horizontal Transverse Axis 19 ( x h) ( y k) = 1 a b The center of the hyperbola is ( h, k ). The foci are ( h ± c, k), where c = a + b. The vertices are ( h ± a, k). A hyperbola can have a vertical transverse axis also as shown next. Transverse Axis The equation for a hyperbola with a vertical transverse axis is as follows. The Standard Form of a Hyperbola with a Vertical Transverse Axis ( y k) ( x h) = 1 a b The center of the hyperbola is ( h, k ). The foci are ( h, k ± c), where c = a + b. The vertices are ( h, k ± a). Example 1: Sketch the graph of the hyperbola. State the coordinates of the center, the vertices and the foci. Find the equations of the asymptotes. ( y + ) ( x 1) = The center is (1, ). The hyperbola has a vertical transverse axis. We will draw the fundamental rectangle using a =, and b = 4. Then sketch the asymptotes and the hyperbola opening up and down. The vertices occur at (1, ± ), or V 1 (1, 0) and V (1, 4). To get the foci we need to use c = a + b, so c = () + (4) = 0 and 007 Linn-Benton Community College MTH 11 Conics

20 0 c = 0 = 5. The coordinates of the foci are (1, ± 5), or F 1 (1, + 5 ) and F (1, 5 ). 4 F 1 V C -4-6 V 1 F To find the equations of the asymptotes, notice that each asymptote passes through the center, (1, ), and a corner of the fundamental rectangle. The positively-sloped asymptote has a rise of and a run of 4, for a slope of = 1. Similarly, the 4 1 negatively-sloped asymptote has a slope of =. Using the point-slope form of 4 the equation of a line we obtain the equations of the asymptotes. y y = m( x x ) y y = m( x x ) y ( ) = ( x 1) y ( ) = ( x 1) x 3 x 5 y = y = ou may have encountered hyperbolas without realizing it. A light bulb in a cylindrical lampshade will cast light on the wall in the shape of a hyperbola. Two pebbles dropped into still water will produce waves in concentric circles, the intersections of which form hyperbolas. Nuclear reactors must have cooling towers that can withstand high winds and be made of as little material as possible. The most efficient shape has a hyperbolic cross section. -8 Lamp Light Intersecting Waves Cooling Tower 007 Linn-Benton Community College MTH 11 Conics

21 Example : Complete the square to put the equation for the hyperbola in standard form. Then state the center and direction of the transverse axis. 1 x 9y 4x 54y =. Example 3: Rearrange the terms to gather x-terms together and y-terms together. Factor so that the leading coefficient on each squared term is 1. x 9y + 4x 54y 97 = 0 ( x + x ) 9( y + 6 y ) = 97 ( + + 1) 9( ) = x x y y ( 1) 9( 3) 18 x + y + = ( x + 1) 9( y + 3) 18 = ( x + 1) ( y + 3) = 1 9 The center of the hyperbola is ( 1, 3) and the transverse axis is horizontal. A cooling tower with a cross section in the shape of a hyperbola is 500 feet tall. Its narrowest point is 375 feet above the ground. If the hyperbola is placed in a coordinate system, with its base centered along the x-axis, then the hyperbola has the following equation. ( y ) x 375 = How wide is the tower at its base? Consider the graph of the hyperbola. The center is at (0, 375). To determine the width at the base, find the x-coordinates when y = 0. The distance between those two points is the width. Substitute y = 0 into the equation and solve for x. ( 0 375) x = x 375 = x = ± x ± The endpoints of the base occur at ( 170.3, 0) and (170.3, 0), so the base has a width of (170.3) = feet Linn-Benton Community College MTH 11 Conics

22 EERCISES: Exercises 1 through 6: Sketch a graph of each of the following hyperbolas, with the fundamental rectangle and asymptotes. State the coordinates of the center, vertices and foci, and give the equations of the asymptotes. x y 1. = diameter of the observation deck. Round to the nearest tenth y x = ( y + 4) ( x + 1) = ( x ) ( y + 3) = ( y ) 49 = ( x 1) 1 Kobe Port Tower 1. Find the diameter of the base of the Kobe Port Tower from Exercise 11. Round to the nearest tenth. 13. A cooling tower, like the one pictured in Example 3, is 450 feet tall. The narrowest point is 390 feet above the ground, where it is 180 feet wide. The base is 30 feet wide. Find the diameter of the top of the tower. Round to the nearest tenth. 6. ( y 3) ( x + 1) = 1 4 Exercises 7 through 10: Complete the square to put the equation for the hyperbola in standard form. State the coordinates of the center and the direction of the transverse axis. 7. 9x y 36x 6y + 18 = y x x = 0 3x y 6x 1y 7 = y x + x + 54y + 6 = The Kobe Port Tower is a hyperbolic telecommunications tower in Kobe, Japan. It has a cross section that is a hyperbola. The height of the tower is 108 meters and there is an observation deck at a height of 90.8 meters. If an equation of the cross section of the tower is x ( y 65.5) = 1, then find the A cooling tower, like the one pictured in Example 3, is 300 feet tall. The narrowest point is 65 feet above the ground, where it is 170 feet wide. The diameter at the top of the tower is 190 feet. Find the diameter of the base of the tower. Round to the nearest tenth. Exercises 15 through 0: Classify the graph of the equation as a circle, a parabola, an ellipse or a hyperbola. 15. x + 4y 6x + 16y + 1 = x y 4x 3 = 0 y y x 4 4 = 0 5x 10x 00y 119 = x + 9y 36x + 6y + 34 = 0 0. x( x y) = y(3 y x) 007 Linn-Benton Community College MTH 11 Conics