Geometry: Conic Sections

Transcription

1 Conic Sections Introduction When a right circular cone is intersected by a plane, as in figure 1 below, a family of four types of curves results. Because of their relationship to the cone, they are called conic sections. The study of conic sections explores an area of mathematics called analytic geometry the application of the skills of algebra to geometric shapes. Apollonius of Perga (c BC ), was the first comprehensive look at conics, defining circle, ellipse, hyperbola and parabola; the four curves or shapes formed by passing a plane through a cone [as in figure 1]. Later work by Omar Khayyam, René Descartes and others helped to further define conics to become useful in solving problems with area and volume as well as solutions to cubic equations and advances in optics, astronomy, cartography and other practical sciences.* In the following four chapters we will study the equations and graphs of the four conic sections and in Chapter 5, we will look at the relations among the equations and the intersections of their graphs. * Rooney, Anne, The Story of Mathematics 008. Page 1 of 0

2 Chapter 1: The Circle When a plane that is parallel to the base of the cone is as in figure, the curve formed by the surface of the cone is a circle. The formal definition of a circle is the set of points at an equal distance [r] from a given point (h, k). Equation of a circle with Center (0, 0) We use the distance formula: d = ( x x ) + ( y y ) 1 1 to write the equation of a circle, first with its center at the origin (0, 0) and radius r. y Let P(x, y) be any point on a circle with center at A(0, 0). Then the distance from point A [the center of the circle] to point P is given by the formula: r A AP = r = ( x 0) + ( y 0) (0, 0) P (x, y) x r = x + y (r) = ( ( x 0) + ( y 0) ) The standard form of the equation of a circle with center at the origin (0, 0) and radius r is x + y = r Example 1.1: Write an equation of a circle with a radius of 5 whose center is at the origin. Graph the circle. Using the standard form for the equation of a circle: x + y = r replace r with 5 (0, 5) x + y = (5) x + y = 5 This is equation of the circle. ( 5, 0) (5, 0) (0, 5) Page of 0

3 Example 1.: Find the radius of a circle whose center is at the origin and whose equation is x + y = 9. Graph the circle. Compare the given equation to the standard form of a circle with center at the origin. x + y = r x + y = 9 (0, 3) Therefore, r = 9, and r = 3 ( 3, 0) (3, 0) Starting a (0, 0), go 3 in all directions. (0, 3) Exercise 1a: I. Write the equation in standard form of each circle with its center at the origin and the given radius. 1. r =. r = 6 3. r = 7. r = 5 5. r = r = r = r =.5 9. r = r = r = r = 5 II. Find the center and radius for each circle with the given equation and graph each circle. 13. x + y = x + y = x + y = x + y = x + y 1 = x 9 + y = x = 6 y 0. 1 y = x 1. x + y = 16. 5x = 15 5y 3. 3x + 3y = x = 6y Page 3 of 0

4 Equation of a circle with Center (h, k) Suppose the graph of the circle x + y = r y is shifted horizontally h units. Then, using the distance formula we see that r = (x h) + y. (0, 0) (h, 0) r x Similarly, a vertical shift of k units (not shown) would result in the equation: r = x + (y k). Thus, for any circle with radius r, by translating the y circle with center (0, 0) to a center (h, k) through a r horizontal and vertical shift, the resulting equation becomes: (x h) + (y k) = r (0, 0) (h, k) x Example 1.3: Find the center and radius of a circle with equation (x + ) + (y 3) = 16. Compare the equation of this circle to the equation of the circle in standard form: (x h) + (y k) = r (x + ) + (y 3) = 16 Therefore h =, since x ( ) = x + ; k = 3 and r = 16, so r = For this equation, center at (, 3) and radius. Example 1.: Find the equation of the circle with center (1, ) and radius 3. Since the (h, k) is the point (1, ), h = 1 and k =. Also, r = 3, so r = 9. Replacing these values in the standard form of the equation of a circle (x h) + (y k) = r We get (x 1) + (y ) = 9 Page of 0

5 Exercise 1b: I. Find the center and radius of the circle with the given equation. Graph the circle. 1. (x 7) + (y 5) = 1. (x ) + (y ) = (x + 3) + (y + 1) = 9. (x + 6) + (y + 8) = 9 5. (x + 5) + (y ) = 1 6. (x 3) + (y + 3) = x + (y 3) = (x 1) + y = (x + ) + y = 10. x + (y + 3) = 18 II. Write an equation in standard form of the circle with the given center and radius. 11. (1, ); 5 1. ( 5, 6); 13. ( 1, 0); 1. (0, 3); ( 3, 5); 16. (1, ); (, 13 ); ( 10, 1); (11, 3 ); 5 0. (5, 3); 1 Completing the square to obtain the standard form of the equation of a circle When the equation of a circle is given in general form: Ax + By + Cx + Dy + E = 0, where A = B, it is helpful to rewrite it in standard form. For this, you must recall the perfect square trinomial from algebra. ( + Δ) = + Δ + Δ For example: (x + 5) = x + x So (x + 5) = x + 10x + 5 Thus, to rewrite a trinomial as the square of a binomial, consider x 1x + 36 = x x (x 6) = (x 6) Page 5 of 0

6 Example 1.5: Geometry: Conic Sections Find the center and radius of a circle with equation x + y + 6x 8y = 0 x + y + 6x 8y = [move the constant to the other side of equal sign] x + x y y + = + + x + x y y + = (x + 3) + (y ) = 9 [standard equation of a circle] Center: ( 3, ); Radius: 7 Exercise 1c: Rewrite each equation in standard form. Find the center and radius. 1. x + y 1x 6y 3 = 0. x + y + 1x + 16y + 8 = 0 3. x + y + y = 15. x + y x = 8 5. x + 8x + y 1y = x 10x + 5y 50y = 0 7. x + y + 36y = 5 8. x x + y = (x + y + ) = 6y 1x 10. 3x x 1 = y + y Page 6 of 0

7 Chapter : The Ellipse When a plane intersects a cone is as in figure 3, the conic section formed is an ellipse. The shape of an ellipse is that of an elongated circle, and its equation is related to the equation of a circle. Formally, an ellipse is defined as the set of points such that the sum of the distances from two given points, called foci, [Fo si] is constant [length of the major axis]. Figure 3 Equation of an ellipse with Center (0, 0) We can derive the equation of an ellipse by (0, 3) using the distance formula. The ellipse at ( 5, 0) F the right has foci at F 1 (, 0) and F (, 0). 1 F (5, 0) (, 0) (, 0) To derive the equation of this ellipse, we show that F 1 P + F P = 10. (0, 3) P(x, y) F 1 P + F P = 10 ( x ( )) + ( y 0) + ( x + 8x y + ( x ()) + ( y 0) = 10 ( x 8x y = 10 ( x + 8x y = 10 ( x 8x y Squaring both sides, x + 8x y = ( x 8x y + x 8x y 16x 100 = 0 x 5 = 5 ( x 8x y ( x 8x y Squaring both sides again, 16x 00x + 65 = 5(x 8x y ) 5 = 9x + 5y 1 = This is the answer in Standard Form of an Ellipse. 5 9 Page 7 of 0

8 The axes of an ellipse are of unequal length. The longest axis is called the major axis [a] and the shortest axis is called the minor axis [b]. In a circle, both axes [or diameters] are of equal length. Therefore the circle is a special case of the ellipse. The standard form for the ellipse iff its major axis oriented horizontally and center at the origin is x a y 1 with a 0, b 0 and a > b + = ; b (0, b) a b F 1 a F ( a, 0) c (a, 0) ( c, 0) (0, 0) (c, 0) (0, b) The foci of an ellipse always falls on the major axis with c = a b. c is the distance from the center to the focus and a is always on the major axis [major radius] and is one-half the length of the major axis, while b is onehalf the length of the minor axis [minor radius]. (0, a) When the major axis of the ellipse, with center at the origin, is vertical the foci fall at (0, c) and (0, c) but you still use c = a b. The standard form of this ellipse is with a 0, b 0 and a > b x b y 1 + = ; a F (0, c) a c ( b, 0) b (0, 0) (b, 0) a F 1 (0, c) Example.1: (0, a) Find the equation of an ellipse with its center at the origin, foci at ( 3, 0) and (3, 0), and the length of the major axis is 1. First we know a = 1 [major axis is always a], so a = 6. Also this is a horizontal ellipse due to the foci are horizontal from each other and that c = 3. Using c = a b find b. (3) = (6) b b = 5, so b = ±5, and the equation for this ellipse is 36 5 Page 8 of 0

9 Example.: Write the equation of the ellipse 9x + y = 36 in standard form. Find its foci and the lengths of the major axis and minor, finally sketch the ellipse. To put the equation in standard from, it must equal 1, so divide both sides of the equal sign by 36: 9x y 36 + = This is now in standard form: 36 Center is at (0, 0) and this is a vertical ellipse [the major axis is vertical] a = 36 and b =, so a = ±6 and b = ±, and c = 36, so c = ± 3 Foci: (0, ), (0, ), Major axis (a): 1 Minor axis (b): (0, 6) F (0, 5.7) (, 0) (, 0) (0, 0) (0, 5.7) F 1 (0, 6) Exercise a: I. Find the center, foci, lengths of the major and minor axes, and sketch the graph of each of the following ellipses II. Write each of the following in standard form of an ellipse, tell whether the ellipse if horizontal or vertical, find the coordinates of the foci and the lengths of the major and minor axes. 7. x + 5y = x + y = y = 6 x 10. 9x 100 = y 11. 5x + 10y = x + 1y = 1 Page 9 of 0

10 Equation of an ellipse with its center at (h, k) When the center of an ellipse falls at (h, k) instead of (0, 0), the standard forms of ellipses become either: ( x h) ( y k) if it is a horizontal ellipse, and a b ( x h) ( y k) if it is a vertical ellipse. b a The values of a, b and c are defined in exactly the same way as for the ellipse centered at the origin. a is the distance from the center to the vertices on the major axis. b is the distance from the center to the vertices on the minor axis. c is the distance from the center to the focus on the major axis. The relationship between a, b and c remains c = a b Example.3: Find the center, the coordinates of the vertices and sketch the graph of the ellipse whose equation is ( x 3) ( y+ 5) 16 1 The center of this ellipse is (3, 5). Since the larger denominator is under the x, the ellipse is horizontal. Since a = 16, a = ± so the horizontal vertices fall units to the right and left of the center: (7, 5) & ( 1, 5). Similarly, if b = 1, then b = ± 3, so the vertical vertices fall 3 units above and below the center: (3, 5+ 3 ) & (3, 5 3 ). Use decimal forms to graph (3, 1.5) & (3, 8.5) Additionally, c = 16 1, c = ±, therefore the foci for example.3 would be on the horizontal axis [right & left of the center] at (5, 5) & (1, 5). Page 10 of 0

11 Example. Find the equation of an ellipse with its center at (1, ) if its major axis is vertical and 10 units long and one focus falls at (1, 1). Sketch the ellipse. Since c is the distance from (1, ) to (1, 1), then c = ±3 and therefore the other focus falls at (1, 5). The major axis a = 10, a = 5, so the vertical vertices are (1, 3) and (1, 7). c = a b 9 = 5 b, so b = 16, b = ±. Hence the horizontal vertices fall units to the right and left of (1, ) at (5, ) and ( 3, ). Replacing the values (h, k) = (1, ), a = ±5, b = ± in the general form for the vertical ellipse, We get Exercise b: ( x h) ( y k) b a ( x 1) ( y+ ) I. Find the center, the coordinates of the vertices and sketch the graph. 1. ( x ) ( y 1). 9 ( x 5) ( y ) 3. 5 x ( y+ ) 9 9. ( x+ 3) y ( x+ 6) ( y ) ( x+ 7) ( y+ 5) 1 10 II. Find the equation of the ellipse with the information given. 7. Center at (, 5) 8. Center at (, 1) major axis vertical. foci at ( 7, 1) and ( 1, 1) a = 5, b =. length of major axis 1 9. Center at (3, ) 10. Vertical ellipse major axis horizontal. vertices of the major axis length of major axis: 10. are at (, 5) & (, 0) length of minor axis:. length of minor axis: Page 11 of 0

12 Completing the square to obtain the standard equation of an ellipse Example.5 Find the center and foci of the ellipse whose equation is x + 8x + 3y 18y = 19 We must complete the square to put the equation in standard form. x + 8x + 3y 18y = 19 (x + x ) + 3(y 6y ) = 19 (x + x + (1) ) + 3(y 6y + (3) ) = 19 + (1) + 3(9) [distributive prop.] (x + 1) + 3(y 3) = 1 Divide both sides by 1 ( x+ 1) ( y 3) + = 3 c = a b c = 3, c = ±1; This is a vertical ellipse whose center is at ( 1, 3) and foci ( 1, ) & ( 1, ) 1 Exercise c: I. Write each in standard form of an ellipse then sketch their graphs. 1. 9x + y y = 0. x + 1x + y = 3. x 0x + y + 6y = x + 6x + 5y 150y = 111 II. Find the center and foci of each of the following ellipses. 5. x 1x + 3y 6y = x + 30x + y + 8y = x + y 1x 8y = 1 8. x + 3y + x + y = x 100x + y + 1y = x 10x + y + y = 3 Page 1 of 0

13 Chapter 3: The Hyperbola Another conic section is the hyperbola [hī olah perb ], which is formed when the cutting plane is parallel to the axis of the cone and cuts both elements of the cone, as in figure. Definition: A hyperbola is the set of points in a plane such that for each point, the difference of its distances from two fixed points [foci] is constant. The hyperbola at the right has foci at F 1 (0, 5) & F (0, 5) and a difference of the distances from the foci to a point on the hyperbola of 8. Let P(x, y) be a point on the hyperbola. You can use the distance formula to find the equation of this hyperbola. This makes the vertices of this hyperbola at V 1 (0, ) and V (0, ). P (x, y) 5 F V 5 5 V 1 5 F 1 F 1 P F P = 8 ( x 0) ( 5) + y+ ( x 0) ( 5) + y = 8 x + ( y+ 5) = 8 + x + ( y 5) Square both sides x + (y + 5) = x + y + 10y + 5 = x x + ( y 5) + x + (y 5) + ( y 5) + x + y 10y + 5 0y 6 = 16 5y 16 = x x + ( y 5) + ( y 5) Square both sides again 5y 160y + 6 = 16[x + (y 5) ] 5y 160y + 6 = 16x + 16y 160y y 16x = 1 Now set equal to 1. Page 13 of 0

14 So the equation for the hyperbola with foci at (0, ±5) and a difference of the distances from the foci to a point on the hyperbola of 8 is y x 16 9 The vertices are where the hyperbola intersects the transverse axis. The transverse axis is the segment between the vertices. The length of the transverse axis is a. The segment perpendicular to this is the conjugate axis and has a length of b. The asymptotes of a hyperbola are lines that the hyperbola approaches as the asymptote goes out from the center of the hyperbola, but never intersects with the hyperbola. The hyperbola and the asymptotes become infinitely close to each other. F V V 1 F 1 Transverse axis (a) Conjugate axis (b) Equation of a Hyperbola with Center (0, 0) y x is the Standard form of the equation of a a b hyperbola with center at the origin (0, 0) and foci at F 1 (0, c) & F (0, c), with a 0, b 0, c = a + b and the equation of the asymptotes is y = y > 0 the transverse axis will be vertical [the hyperbola opens up & down]. ± a b x. When F V V 1 F 1 a b = 1 is the Standard form of the equation of a hyperbola with center at the origin (0, 0) and foci at F 1 ( c, 0) & F (c, 0), with a 0, b 0, F V V 1 F 1 c = a + b and the equation of the asymptotes is y = ± b a x. Slope is always y x. When x > 0 the transverse axis will be horizontal [hyperbola opens up & down]. Page 1 of 0

15 Example 3.1: Draw the graph of 16 The center is at (0, 0) and since x is positive [in front] then the transverse axis will be horizontal [left & right]. a = 16, a = ±, so the vertices will be (±, 0) and b =, b = ±. The equation of the asymptotes will be y = ± 1 x The asymptotes are dotted lines because they should be erased after you have drawn the curves. Example 3.: Draw the graph of y x 36 9 The center is at (0, 0) and since y is positive [in front] then the transverse axis will be vertical [up & down]. a = 36, a = ±6, so the vertices will be (0, ±6) and b = 9, b = ±3. The equation of the asymptotes will be y = ±x Exercise 3a: Draw the graphs of the following y x y x x y y x ( x ) ( y+ 1) * 9 16 *use the same rule for center as an ellipse. Page 15 of 0

16 Example 3.3: Find the vertices, foci and equation of the asymptotes of the hyperbola and 5x y = 100. Draw the graph. 5x y 100 = in standard form. 5 The center is at (0, 0) and the transverse axis is horizontal. a = ± so the vertices are at (±, 0) and b = ±5. The foci are found using c = a + b c = + 5 c = 9 c = ± 9 5. So the foci are (± 9, 0), 5 F 1 V 1 V F 5 and the asymptotes are y = ± 5 x Exercise 3b: I. Find the vertices, foci, and equations of the asymptotes for each of the following hyperbola. Draw the graph y x y x 5 11 II. Write the following hyperbola in standard form, then find their vertices, foci, and equations of the asymptotes. 5. 5y x = 5 6. x y = y 9x = 1 8. x 16y = y 5x = x 9y = x = 1 + 3y 1. y y x + 10x = 5* *use the same rule as an ellipse. Page 16 of 0

17 Example 3.: Write the hyperbola in standard form that has vertices are (0, ±3 asymptotes of y = ±3x. 10 ) and c = ± 90, and if (0, ±c) then the transverse axis is vertical Asymptotes are y = ± a b x and a b = 3, so a = 3b Foci: c = a + b 90 = (3b) + b 90 = 10b 9 = b, so b = 3 and a = 9 y x Filling in the equation a b y x 81 9 Exercise 3c: Write the equation in standard form, for each hyperbola described below. 1. Vertices: (±, 0). Vertices: (0, ±) Asymptotes: y = ± 3 x. Asymptotes: y = ± 5 x 3. Foci: (0, ±3 5 ). Foci: (±5, 0) Asymptotes: y = ±x. Asymptotes: y = ± 3 x 5. Vertices: (±3, 0) 6. Vertices: (0, ±9) Foci: (± 3, 0). Foci: (0, ± 106 ) 7. Vertices: (0, ±) 8. Vertices: (±, 0) Foci: (0, ± 5 ). Foci: (± 9, 0) Page 17 of 0

18 Chapter : The Parabola When the cutting plane that is parallel to one of the elements of the cone is passed through the cone as in figure 5 to the right, the curve formed is a parabola. The graph of the quadratic function, y = ax + bx + c, where a, b, and c are real numbers, and a 0, is the parabola. Equation of a Parabola with Vertex [Center] at (0, 0) Let us look first at the graph of the equation where a = 1 and b & c = 0 (y = x ). By making a partial table of values we get the set of points as follow: x y Plotting these points in the xy plane we get: Notice that for every point (x, y) on the parabola, there is a corresponding point ( x, y). This is called the symmetry of the parabola; the axis of symmetry is the line x = 0 (for this parabola). There is one point of the parabola which intersects the line of symmetry, and is the point where the parabola turns. This is called the vertex of the parabola. For this equation, the vertex is (0, 0). Page 18 of 0

19 Next, let s look at equations of the form y = ax, ie. a is any real number except 0 while b = 0 and c = 0. For example, consider the graphs of the equations: a: y = 1 x, b: y = x, c: y = x First, make a table of values for each equation. a. b. c. x 1 x x x x x Plotting these on the same set of axes, we get: By observing these we can conclude, for y = ax, the vertex is (0, 0), the axis of symmetry is x = 0. If a > 0 [positive], the parabola opens upward, and * a b if a < 0 [negative], it opens downward. Similarly, for the graph of relation x = ay, we get the vertex is (0, 0), axis of symmetry, y = 0. If a > 0 the parabola opens to the right, and if a < 0 the direction of opening is to the left. c * Dotted parabola: y = x Page 19 of 0

20 Example.1: For the parabola, state the vertex, axis of symmetry and direction of opening, then sketch the graph for x = 3y. Solve for x [the non-squared variable]: x = 3 y. Now make a table of values. y y Vertex: (0, 0) Axis of symmetry: y = 0 (horizontal) Parabola opens to the left (a < 0) Exercise a: For each parabola, sketch the graph, include the vertex and axis of symmetry. 1. y = 1 3 x. y = 3x 3. x = y. x = y 5. y = x 6. y = x 7. y + 3x = 0 8. x y = 0 Page 0 of 0

21 Parabola with vertex not at the origin Now, consider the graphs of y = x + 3 and y = (x + 3) 6 8 Axis of Symmetry Axis of Symmetry Vertex: (0, 3). Vertex: ( 3, 0) Axis of symmetry: x = 0 (vertical). Axis of symmetry: x = 3 (vertical) Parabola opens downward. Parabola opens upward From this we see the equation of the form y = ax + k shifts the vertex vertically on the graph and the form y = a(x h) shifts the vertex horizontally. Combining them, we get an equation of the form y = a(x h) + k Standard Equation of a Parabola axis of equation vertex symmetry direction of opening y = a(x h) + k (h, k) x = h a > 0: upward a < 0: downward x = a(y k) + h (h, k) y = k a > 0: to the right a < 0: to the left Example.: State the vertex, axis of symmetry, and direction of opening, then sketch the graph. x = (y + 3) + 3 Page 1 of 0

22 Now make a table of values. y ( y 3) This is getting too big, try going smaller Vertex: (3, 3) Axis of symmetry: y = 3 (horizontal) Parabola opens to the right Axis of Symmetry Exercise b: I. For each parabola, sketch the graph include the vertex and axis of symmetry. 1. y = (x 1) + 3. y = (x + 1) + 3. x = 1 3 (y 3) 5. x = (y + ) II. State the vertex, axis of symmetry, and direction of opening. 5. y = x y = 3(x + 1) 6 7. x = (y 1) x = (y + ) x = y y = 1 3 (x ) + 3 Completing the square to obtain the standard equation of a parabola If the quadratic equation is written in the general form y = ax + bx + c, it is useful to rewrite it in standard form of a parabola [y = a(x h) + k]. This is done by completing the square. Example.3: Express the equation y = x 8x 5 in standard form of a parabola. y + 5 = x 8x y () = x 8x + () y + 1 = (x ).y = (x ) 1 Vertex (, 1), Axis: y = 1, parabola opens upward Page of 0

23 Example.: Express the equation x = y + 10y + 9 in standard form. x 9 = (y + 5y ) x 9 + [ 5 ] = (y + 5y + [ 5 ] ) x = (y + 5 ) x = (y + 5 ) 7 Vertex ( 7, 5 ), Axis: x = 7, parabola opens to the right Exercise c: I. Rewrite each equation in Standard Form of a Parabola, then sketch the graph. 1. 6y = x. y = 1 8 x 3. (y ) = 3x. (x + 3) = 1 y 5. (y + 1) + (x + 5) = 0 6. y = x 6x + 5 II. Rewrite each equation in Standard Form of a Parabola, then state the vertex, axis of symmetry, and direction of opening. 7. x = y + 8y y = x 10x y = x + 6x x = y + 11y 11. y = x x 1 1. y = 3x + 6x y = 3x x x = y + 8y x = 3y + y + 5 Focus and Directrix of a parabola A parabola is defined as the set of all points equidistant from a fixed line called the directrix and a fixed point, not on the line, called a focus. The vertex and focus are both on the axis of symmetry. The directrix is perpendicular to the axis of symmetry. The distance from the vertex to the focus equals the distance from the vertex to the directrix. That distance is called p. In the standard form y a(x h) + k, the distance of p = 1 a. Directrix Axis of Symmetry F p V p Page 3 of 0

24 The distance from the focus to the intersection of the directrix and axis of symmetry is p. If the vertex is (h, k), then the focus is (h, k+ 1 ) and the directrix is y = k 1 a Similarly, for the standard form x = a(y k) + h [parabola that opens right or left], then the focus is (h + 1, k) and the directrix is x = h 1 a. a a Example.5: Name the vertex, axis of symmetry, focus and directrix for y = (x 1) + Parabola opens up, with a = Vertex: (1, ) Axis of symmetry: x = 1 Focus: p = 1 () = 1 8, so (1, ),. or (1, 17 8 ) 1 (1,.15) F Directrix: V y = (1, ) Directrix: y = 1 8, or.y = [directrix is a line] Example.6: Name the vertex, axis of symmetry, focus, and directrix for x = 1 (y + 3) 1 Parabola opens to the left Vertex: ( 1, 3) Axis of symmetry:.y = 3 Focus: p = 1 ( 1) = 1, so ( 1 1, 3),.. or ( 3, 3) Directrix: x = 1+ 1, or.x = ( 1.5, 3) F V ( 1, 3) 3 5 Directrix x = 0.5 Page of 0

25 Exercise d: I. Sketch the graph for each parabola, include the vertex, focus and directrix [clearly label the coordinates or equations for each]. 1. y = 1 x. x = 3 y + 3. x = (y + ). y = 1 8 (x 1) + 1 II. Find the coordinates of the vertex, focus and the equation of the directrix. 5. y = 1 10 x 6. y = 1 (x ) x = 1 6 y x = 8(y ) y = 3(x + ) y = 3 (x 1) 3 The equation of a parabola using the focus and directrix. Given the coordinates of the focus and the equation of the directrix, it is possible to determine the equation of the parabola. Example.7: Write an equation in standard form of a parabola given the Focus (, ) and directrix x = 1. Let (x, y) be a point on the parabola, then the distance from (x, y) to the focus equals the distance from (x, y) to the directrix. Using the distance formula [ d = ( x x ) + ( y y ) ] we get: 1 1 ( x 1) ( y y) + = ( x ) + ( y ) Square both sides: (x 1) + (0) = (x ) + (y ) x x + 1 = x x + + (y ) x = (y ) D(1, y).. F(, ). P(x, y) x = 1 (y ) Page 5 of 0

26 Exercise e: I. Write the equation in standard form of a parabola. 1. F:(0, 3); Directrix: y = 3. F:(7, 0); Directrix: x = 1 3. F:(, ); Directrix: x = 0. F:(, ); Directrix: y = 0 5. F:(0, 0); Directrix: y = 1 6. F:(0, ); Directrix: x = 1 II. For each parabola, sketch the graph and label the vertex, then write the equation. 7. F:(0, ); Directrix: y = 8. F:(1, ); Directrix: y = 0 9. F:( 1, 0); Directrix: x = F:(1, ); Directrix: x = 0 Page 6 of 0

27 Chapter 5: Equations for the Conic Sections We have now studied some of the basic information about conic sections and their associated equations. In addition to the fact that these curves are generated when a plane intersects a cone is the fact that the equations for the conic sections are related by any equation of the form Ax + By + Cx + Dy + E = 0 The graph of an equation of this form is a conic section. More specifically, the four we have just completed can be described as follows: 1. If A = B and A 0, then the graph is a circle.. If A and B have the same sign [non-zero], then the graph is an ellipse. 3. If and A and B have different signs, then the graph is a hyperbola.. If either A or B are zero [not both], then the graph is a parabola. Example 5.1: Identify the following equation as a Circle, Ellipse, Hyperbola, or Parabola. a. (x ) + (y ) = (x 1) b. 3x = x 3y put into standard form for all conics [Ax + By + Cx + Dy + E = 0] x x + + y y + = x + x + 1 y 6x y + 7 = 0. 3x + 3y x = 0 A = 0, B = 1 [#]. A = 3, B = 3 [#1] Parabola. Circle Exercise 5a: Identify the following equation as a Circle, Ellipse, Hyperbola, or Parabola. 1. x + y = 16. y = (x 5) y x x = (y + 7) ( y ) ( x+ ) x = 9 y 8. x y = x + y x + 6y + 1 = x + y = x = y 1x (x + ) + 9(y 3) = y = x 1. x 1y + y = x + 13y 6x = 8 Page 7 of 0

28 Conic Section Review CIRCLE: General Form: Ax + Ay + Cx + Dy + E = 0 Standard Form: (x h) + (y k) = r Center: (h, k) Radius: r y C (h, k) P (x, y) r x ELLIPSE: General Form: Ax + By + Cx + Dy + E = 0 Horizontal Major Axis Vertical Major Axis F C F F C F ( x h) ( y k) ( x h) ( y k) Standard Form: a b b a (h, k) Center: (h, k) (h±a, k) horizontal vertices (h±b, k) (h, k±b) vertical vertices (h, k±a) (h±c, k) Foci (h, k±c) a major axis length a b minor axis length b c sum of the focal radii c c = a b Page 8 of 0

29 HYPERBOLA: General Form: Ax By + Cx + Dy + E = 0 Horizontal Transverse Axis Vertical Transverse Axis F V F V V 1 F 1 V 1 F 1 ( x h) ( y k) ( y k) ( x h) Standard Form: a b a b (h, k) Center: (h, k) (h±a, k) vertices (h, k±a) (h, k±b) conjugate axis endpoints (h±b, k) (h±c, k) Foci (h, k±c) y = ± b a x Equation of Asymptotes y = ± b a x a Transverse axis length a b Conjugate axis length b a Difference of the focal radii a c = a + b PARABOLA: Opens Vertically (Function) Opens Horizontally F V Directrix Axis of V F Symmetry Axis of Symmetry Ax + Cx + Dy + E = 0 General Form By + Cx + Dy + E = 0 y = a(x h) + k Standard Form: x = a(y k) + h (h, k) Vertex: (h, k) (h, k+ a 1 ) Focus (h+ 1 a, k) y = k 1 a Directrix x = h 1 a a > 0: Up; a < 0: Down Parabola opens: a > 0: Right; a < 0: Left Directrix Page 9 of 0

30 Exercise 5b: 1. Find the equation of a circle that passes through the origin and has its center at (5, 1).. A circle has a diameter with endpoints ( 7, 5) and (1, 1). Find the equation of the circle. 3. P(1, ), Q(, 3), and R(5, 0) are vertices of a triangle that is inscribed in a circle. Find the equation of the circle.. A pool table is shaped as an ellipse, with its major axis 18 feet long and its minor axis 1 feet long. If a cue ball places at one focus is struck so that it bounces off one cushion and stops at the other focus point, how far has the cue ball traveled? 5. The grass area in front of the local city hall is in the shape of an ellipse, 0 feet wide and 100 feet long. There is a flag pole at each focus. How far apart are the flag poles? 6. The ellipse with equation 5x + 16y 00 = 0 is moved 5 units to the right and 3 units down. Find the equation of the ellipse in its new position. 7. Graph ( x+ 1) ( y+ ) 16 9 = 1. Hint: use the basic graph for Write the equation y x 6y + 3x 71 = 0 in standard form for a hyperbola which does not have its center at its origin. 9. Draw the graph of xy = 6, then give the equation of the asymptotes [Hint: solve for y, then make a table of values to plot points]. 10. Find the equation of the parabola which has points at (0, 7) & (, 5) and the axis of symmetry is x =. 11. In a suspension bridge the horizontal distance between the supports is the span of the bridge, and the vertical distance between the points where the cable is attached to the supports and the lowest point of the cable is the sag. The span of the George Washington Bridge is 3500 feet, and its sag is 316 feet. Find an equation of the parabola that represents the cable. 1. The center of gravity of a long jumper during a given jump traces a curve given by y = 5 1 x x + 3 where y is the height of center of gravity at each point of x [the length of the jump]. Determine the maximum height of the center of gravity during the jump. Hint: complete the square to put into standard form, don t forget to distribute the negative. Page 30 of 0

31 p.3 Ex.1a Section I Geometry: Conic Sections ANSWERS 1] x + y = 16 3] x + y = 9 5] x + y = 1 1 7] x + y = ] x + y = 1 16 p.3 Ex.1a Section II 13] C: (0, 0); 15] C: (0, 0); r =. r = ] x + y = 13 17] C: (0, 0); * 19] C: (0, 0); 8 r = 1 r = 8 8 1] C: (0, 0); 3] C: (0, 0); r = 9. r = 7.6 p.5 Ex.1b Section I 1] C: (7, 5); 3] C: ( 3, 1); r = 1. r = 7 5] C: ( 5, ); 7] C: (0, 3); r = 1.6. r = 6. 9] C: (, 0); r = 6.9 Page 31 of 0

32 p.5: Ex.1b Sec. II 11] (x 1) + (y ) = 5 13] (x + 1) + y = 15] (x + 3 ) + (y 5) = 16 17] (x ) + (y + 13 ) = 1 19] (x 11) + (y 3 ) = 0 p.6 Ex.1c 1] (x 7) + (y 3) = 81; Center: (7, 3); radius: 9 3] x + (y + 1) = 16; Center: (0, 1); radius: 5] (x + ) + (y 3) = 13; Center: (, 3); radius: ] x + (y + 9 ) = 19; Center: (0,.5); radius: 19. 9] (x + ) + (y 1) = 1; Center: (, 1); radius: 1 Page 3 of 0

33 p.9 Ex.a Sec. I 1] Center: (0, 0) Foci: (0, ± 5 ) Major axis: 6 minor axis: F ANSWERS F 3] C: (0, 0). F: (±3, 0) Maj. axis: 10 F F min. axis: 8 5] C: (0, 0) F: (0, ±7) F Maj. axis: 7 Min. axis: 10 F p.9 Ex.a Sec. II 7] 9] 11] Horizontal. Horizontal. Vertical Foci: (± 1, 0). Foci: (± 3, 0). Foci: (0, ± 6 ) Maj. axis: 10. Maj. axis: 8;. Maj. axis: Min. axis:. Min. axis:. Min. axis: Page 33 of 0

34 ANSWERS p.11 Ex. b Sec. I: 1] C: (, 1) 3] C: (0, ) (, ). (7, ), (, ) 1 C. ( 7, ) (, 1).. (0, 1) (0, 1). (0, 5) C 5] C: ( 6, ) ( 6 ±, ) or ( 3., ), ( 8.8, ) ( 6, 6), ( 6, ) p.11 Ex. b Sec. II: C ] ( x ) ( y 5) 9] 16 5 ( x 3) ( y+ ) 5 p.1 Ex. c Sec. I: 1] x ( y 3) 3] 9 6 ( x 5) ( y+ 3) 1 C 6 6 C p.1 Ex. c Sec. II: 5]. 7] 9] ( x 3) ( y 1) ; Center: (3, 1); Foci: (, 1), (, 1) 3 ( x 1) ( y 1) ; Center: (1, 1); Foci: (1, 1± ) 6 ( x ) ( y+ 6) ; Center: (, 6); Foci: (, 6± 6 ) 1 5 Page 3 of 0

35 ANSWERS p.15 Ex.3a 1] 3] Vertices: (±5, 0) 3. Vertices: (±, 0) y = ± 3 5 x y = ± 3 x 3 3 5] 7] Vertices: (0, ±). Vertices: (±7, 0) y = ± = ± 9 7 x ] Center (, 1) Vertices: ( 1, 0), (5, 0) y = ± 3 x p.16 Ex.3b Sec. I 1] Vertices: (0, ±1); Foci: (0, ±13) Asymptotes: y = ± 1 5 x 13 F F. 3] Vertices: (±, 0);. Foci: (± 6, 0). Asymptotes: y = ± 6 x.9 F F...9 Page 35 of 0

36 p.16 Ex.3b Sec. II ANSWERS 5] y x y x 7] V: (0, ±1). V: (0, ±3) Foci: (0, ± 6 ). Foci: (0, ±5) Asymptotes: y = ± 1 5 x. Asymptotes: y = ± 3 x 9] y x 11] V: (0, ±5) V: (± 3, 0);. Foci: (0, ± 3 ). Foci: (± 7, 0) p.17 Ex.3c. Asymptotes: y = ± 5 x Asymptotes: y = ± x 1] 5] 9 3] 9 5 7] y x 36 9 y x 1 Page 36 of 0

37 ANSWERS p.0 Ex.a Axis of Sym 1] y = 1 3 x x = 0 3] x = y V: (0, 0) V (0, 0) Axis of Sym y = 0 Axis of Sym x = 0 5] y = 1 x 7] x = 1 3 y V: (0, 0) Axis of Sym y = 0 V (0, 0) p. Ex.b Sec.I 1] 5 3] V (1, 3) V ( 5, 3) Axis of Sym y = 3 Axis of Sym: x = 1 8 p. Ex.b Sec. II 5] vertex: (0, 3) 7] vertex: (3, 1) 9] vertex: ( 5, 0) axis of sym: x = 0. axis of sym: y = 1. axis of sym: y = 0 opens down. opens to the left. opens to the right Page 37 of 0

38 ANSWERS p.3 Ex.c Sec. I Axis of Sym: 1] y = 1 6 x x = 0 3] x = 1 (y ) 3 V (0, 0) V (0, ) Axis of Sym y = 5] x = (y + 1) 5 Axis of Sym y = 1 V ( 5, 1) p.3 Ex.c Sec. II 7] x = (y + ) + 9] y = (x + 3) 8 11] y = (x 1 ) 5 vertex: (, ). vertex: ( 3, 8). vertex: ( 1, 5 ) axis of sym: y =. axis of sym: x = 3. axis of sym: x = 1 opens to the right. opens up. opens up 13] y = 3(x + ) 15] x = 3(y ) ; vertex: (, ). vertex: ( 11 1, 1 6 ) axis of sym: x =. axis of sym: y = 1 6 opens down.. opens to the right p.5 Ex.d Sec.I 1] p = 1 3] p = 1 F (0, 1) V (0, 0) Directrix: y = 1 ( 1 Directrix x = 1 F V, ) (0, ) Page 38 of 0

39 p.5 Ex.d Sec. II Geometry: Conic Sections ANSWERS 5] Vertex: (0, 0) 7] Vertex: (1, 0) 9] Vertex: (, 7) p = 1 ( 1 ) 10 = 5. p = 1 ( 16 ) Focus: (0, 1 ). Focus: ( 1 Directrix: y = 1. Directrix: x = 1 1 = 3. p = 1 (3) = 1 1, 0). Focus: (, ). Directrix: y = p.6 Ex.e Sec.I 1] p = 3: y = 1 1 x 3] p = : x = 1 8 (y + ) 5] p = 1 : y = x 1 p.6 Ex.e Sec.II 7] y = 1 8 x 9] x = 1 y F V (0, 0) F V (0, 0) Page 39 of 0

40 ANSWERS p.8 Ex.5a 1] Circle 3] Hyperbola (up&down) 5] Parabola (right) 7] Circle 9] Circle 11] Parabola (up) 13] Hyperbola (up&down) 15] Circle p.30 Ex.5b 1] (x 5) + (y + 1) = 13 r = (5 0) + ( 1 0) 3] (x 3) + (y 1) = 5 right triangle so the diameter is PR (1, ) to (5, 0) 5] 0 1 ft 91.7 ft; = c; c = 100 ; c = ] y C ( 5, ) V V (3, ) ( 1, ) x y = ± 3 x 9] rectangular hyperbola; Asymptotes: x & y Axes y = 6 x : 1 1 x y und ] y = x or y = (x 1750) x ; y = a(x 1750) 316 then use (0, 0) Page 0 of 0