# Conic Sections. 1. The type of graph can be determined by looking at the terms.

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1 Conic Sections General Quadratic Equation in Two Variables The general quadratic equation in two variables can be written as A + By + Cy + D + Ey + F = 0 where at least one of the variables A, B, C is not zero. In this class, we will only look at those cases where, that is, there is no y term. The presence of an term B = 0 results in a rotated graph and is covered in a Trigonometry Analytic Geometry course. The graph of the general quadratic equation in two variables can be one of nine things. Seven of these things can be fmed slicing a double napped cone with a plane, so they're often called conic sections. There are graphs of these conic sections in your tet. Determining the type of graph 1. The type of graph can be determined by looking at the terms.. If there are any linear terms, then you should befe determining the type of graph. Eample of completing the square. + y 4+ 6y+ 7= = = 6 ( ) ( y y ) ( ) ( y ) Squared terms are both positive - ellipses circles Here are some eamples in standard fm. + y = 4 y + = ( 3) ( y+ ) + = Here are some eamples that aren't in standard fm, but are still easy to tell the type of graph by inspection y = 8 + y = 1 y

2 Here are two eamples that may provide difficulty because there are linear terms. You really need to complete the square first befe determining the type of graph. y y = 0 y y y = 0 3. If the and are both positive and have the same coefficient, then you have a. y 4. If the and are both positive but have different coefficients, then you have an. y 5. If the and are both positive but the right side is, then the graph is a point. y 6. If the and are both positive but the right side is, then there is no graph. Squared terms have different signs - hyperbola Here are some eamples in standard fm. y = 1 y = ( 3) ( y+ 4) = Here are some eamples that aren't in standard fm, but are still easy to tell the type of graph by inspection. 3y = 4 y = 5 Here are some eamples that you really need to complete the square on befe determining the type of graph. 4 3y + 4 6y= 9 y 5y 4 3+ y 9= 0 7. If the and have different signs, then you have a. y 8. If the and have different signs, but the right side is, then you have intersecting lines.

3 Only one variable is squared - parabolas Here are some eamples in standard fm. = 4y ( y ) = 8( + 3) Here are some eamples that aren't in standard fm, but are still easy to determine what the graph is by inspection y= 9 = 6 y = 4 9. If one variable is squared and the other variable is, then you have a parabola. 10. If one variable is squared and the other variable is, then you have parallel lines. 11. It's possible to also get when only one variable is present. This would be when the solutions are comple numbers involving i. 1. You usually don't need to complete the square to determine the type of graph when only one variable is squared. The question is whether not the second variable is. No variables are squared - lines Eamples of lines are 3+ y= 4 6 = y = If neither variable is squared, then you have a. Practice: Identify the type of graph by inspection. 3 y = = 0 + y = 9 + y = 3 + 3y = 4 4 3y= + 5y = 0 3 4y = y = 8

4 Circles The standard fm f a circle is + y = r 14. The center of the circle is at the. 15. The radius of the circle is. 16. You may shift the circle by replacing the by and the y by. 17. This will move the center to the point (, ) h + y k = r The standard fm f the shifted graph is ( ) ( ) h y k 18. Rather than memizing the standard fm with the and in it, just think about the translations we talked about in chapters 1, 3, and 4. If you see an 3, think and if you see, think. + = y 4 y = Practice: Find the center and radius of the following circles. ( + 1) + ( y ) = 9 ( y ) + 7 = 8 Completing the square Whenever you're wking with circles that have linear terms in them, you're going to have to complete the square to find the center and radius. Consider the circle = 0 y y 19. The first step is to move the to the other side and then the terms together by variable. Leave a space after the linear terms f the net step = 1 y y

5 0. Now complete the square f both the and y terms by taking the linear coefficient and squaring it. Write that in the space you left in the previous step and add it to the other side as well = y y 1. Now, the left side using perfect square trinomials and the right side. ( ) ( y ) = 17. The center of that circle is (, ) and the radius is. Practice: Find the center and radius of = 0 y y Finding the equation of a circle from the graph. 3. The first step is to find out where the is. 4. To do this, identify the codinate of the highest lowest point and the codinate of the points furthest left right. These two codinates give you the center. Now, put a point at the center and find out how far it is to any point on the graph. This is the radius. Finally, write the equation of the circle. Remember to square the radius. The equation of that circle is ( ) 1 + y+ = 4 1

6 5. Another way to find the center is to draw and diameters through the circle. The center is the point where they. Practice: Find the equation of the circles Ellipses The best equation to conceptualize an ellipse is. 6. In this fm, the center is at the. y + = 1 Δ Δy 7. The distance you go from the center in the direction is and the distance you go from the center in the y direction is. 8. If the center isn't at the, then replace the with and the y with y k. That gives you the equation. h y k + = 1 Δ Δy h 9. Notice that everything affecting the hizontal is grouped with the and everything affecting the vertical is grouped with the.

7 Practice. Find the equation of the ellipse. Sketching an Ellipse 30. To sketch an ellipse, start at the. Δ 31. From the center, move units to the and and put dots there. Move units and and put dots there. Δy 3. Draw an ellipse through the four. Practice: Sketch the graph of the ellipses. + 1 y + = ( 1) ( y+ ) + = 1 5 7

8 Standard Fm f Ellipses Hizontal Maj Ais h y k + = 1 a b ( h) ( y k) a + = 1 b Vertical Maj Ais h y k + = 1 b a ( h) ( y k) b + = 1 a 33. The value of a is the of and. The value of b is the Δ Δy of and. 34. The ais is whichever one is longer and the ais is whichever ais is shter. 35. That means that the length of the maj ais is and the length of the min ais is. Δ 36. The are at the end of the maj ais. 37. The direction of maj ais depends on which value,, is. Δy Δ 38. The always lie on the maj ais within the ellipse. 39. The distance from the center to the vertices is. 40. The distance from the center to the endpoints of the min ais is. Δy

9 41. The distance from the center to the foci is. There is a Pythagean relationship between a, b, and c. a = b + c 4. Notice the value goes on a side by itself. Practice: Identify the center, whether the maj ais is hizontal vertical, and find the values of a, b, and c. y + = ( + 3) ( y ) + = To find the codinates of the vertices, start at the and move a units in the direction of the maj ais. If the maj ais is, then add and subtract a to (from) the value. If the maj ais is, then add and subtract a to (from) the y value. Leave the other codinate alone. 44. To find the codinates of the foci, start at the center and move units in the direction of the maj ais. As an eample, if the center is at ( 5,3), the length of maj ais is 8, the focal length is 3, and there is a hizontal maj ais, then we would proceed as follows. 45. Since the length of the maj ais is 8, the value of a is since the maj ais has length of a. 46. Since the maj ais is hizontal, we change the codinates. ( 5± 4,3) becomes ( 5+ 4,3) and ( 5 4,3), which is ( 9,3 ) and ( 1, 3). 47. The foci are units from the center and we once again change the values, so the foci are at 5 ± 3,3. ( )

10 Definition of an Ellipse We've done a lot of wk with Ellipses, but we haven't defined them yet. 48. An ellipse is the set of all points in a plane such that the of the distances from two fied points is constant. 49. Those two fied points are the. 50. The constant is the length of the ais. Completing the square Sometimes, you'll need to complete the square to put the equation of the ellipse into standard fm. You need to be really careful when you do this because now there are coefficients in front of the and. Complete the square and put into standard fm y 8+ 1y+ 4= 0 y 51. Begin by moving the to the other side and the and y terms together y + 1y= 4 5. the coefficient on the out of both 's, even if it doesn't go in evenly. Do the same thing with the y's. Leave space after the linear term but inside the parentheses. ( ) ( y y ) = Take the linear coefficient and square it. Write that value in the spot you left in the previous step. However, remember that there is a constant that you facted out and that what you really just added was the constant times the number you wrote. Add that amount to the other side f both the and the y terms y + 4y + 4 = ( ) ( )

11 54. the left side using perfect square trinomials and the right side. ( ) ( y ) = Finally, divide by the right hand side to make it and put the equation into standard fm. Reduce any fractions so that the entire value is in the denominat. ( 1) ( y+ ) + = The center is at (, ). The change in the direction is and the change in the y direction is. Be careful if you have fractions after you divide. ( ) ( y+ ) = ( 1) ( y+ ) 4/3 + 9/5 = 1 a b Write as instead. You need to do this so that you can figure out what the and are. In this case, a = 9/5 and b = 4/3 (remember a is bigger than b). Take the square roots to get and. a =Δy 1.34 b =Δy 1.15 Practice: Complete the square; find the center, change in and y, and codinates of the vertices and foci y y 156 = 0

12 Hyperbolas 57. The best equation to conceptualize a hyperbola that opens is. y = 1 Δ Δy 58. The best equation to conceptualize a hyperbola that opens is. y = 1 y Δ Δ 59. In either fm, the hyperbola opens in the direction of the variable. 60. In these fm, the center is at the. 61. The distance you go from the center in the direction is and the distance you go from the center in the y direction is. h 6. If the center isn't at the, then replace the with and the y with y k. h y k = 1 Δ Δy That gives you the equation. y k h = 1 y Δ Δ 63. Notice that everything affecting the hizontal is grouped with the and everything affecting the vertical is grouped with the. 64. Instead of having a maj ais and a min ais like an ellipse, the hyperbola has a ais and a ais. 65. The transverse ais is always in the direction of the variable. 66. The conjugate ais is always in the direction of the variable. The conjugate ais reminds us of comple conjugates from imaginary numbers. Likewise, the conjugate ais is imaginary, it's not really there on the graph.

13 Sketching a Hyperbola Sketching a hyperbola starts off like sketching an ellipse. 67. You start at the. 68. From there move units to the left and right and units up and down. 69. Draw a around those four points, so that they are at the center of each side. This is not part of the actual graph, it's just an aid to help us sketch it. 70. Draw dashed lines through the opposite cners of the bo. These make that serve as guidelines f sketching the hyperbola. These are not part of the actual graph, just aids to help us sketch it. 71. The hyperbola touches the bo on the sides of the variable. If the is positive, it will touch on the left and right. If the y is positive, it will touch on the top and the bottom. 7. Sketch the hyperbola, making sure you don't cross the. Practice: Sketch the following hyperbolas. ( ) y y + = 1 ( 1) =

14 Standard Fms Hizontal Transverse Ais h y k = 1 a b ( h) ( y k) a = 1 b Vertical Transverse Ais y k h = 1 a b ( y k) ( h) a = 1 b 73. is distance from the center to the vertices. 74. a is always associated with the variable. 75. The vertices are at the ends of the transverse ais, so the length of the transverse ais is. 76. is the distance from the center to the endpoints of the conjugate ais. Those endpoints don't actually show up on the graph, ecept to help us make the rectangle. 77. b is always associated with the variable. 78. is the distance from the center to the foci. The foci are always inside the curved ption of the hyperbola. 79. Since the are the furthest of the three points from the center, the Pythagean identity f a hyperbola is. c = a + b 80. Notice this is the same identity you're used to seeing from.

15 Practice: Identify the center, whether the transverse ais is hizontal vertical, and find the values of a, b, and c. ( 3.1) ( y+ 1.) = y 3 = To find the codinates of the vertices, start at the and move a units in the direction of the variable. 8. To find the codinates of the foci, start at the center and move units in the direction of the positive variable. ( ) As an eample, if the center is at 4,, a = 3, c = 5, and the y is the positive variable, then we would proceed as follows. 83. Since the is the positive value, we will be changing the value and leaving the value alone. 84. The vertices are units from the center. The codinates of the verte will be at 4, 3. This simplifies to and 4, 1. ( ± ) ( 4,5 ) ( ) 85. The foci are units from the center. The codinates of the foci will be ( 4, ± 5 ). Definition of a Hyperbola 86. A hyperbola is the set of all points in a plane such that the of the distances from two fied points is a constant. 87. Those two fied points are the. 88. The constant is the length of the ais.

16 89. Notice the distance is written with an absolute value because we don't know whether is. d1 d Completing the Square This is very similar to completing the square f an ellipse, ecept one of the variables will have a negative constant facted out of it. Complete the square and put into standard fm: 5y y 16 = Begin by moving the to the right hand side and the and y terms together. 5y 0y 6 + 1= the coefficients on the squared terms out of both terms f each variable. Be sure to fact a negative sign out with the 's in this case. Leave space f another number to go inside the parentheses. ( y y ) ( ) = Take the linear term and square it. Write that in the space you left in the previous step. Multiply the value you added by the constant in front of the parentheses and add this to the other side of the equation. ( y y ) ( ) = Notice in the last step that one of the terms was actually subtracted from both sides. Now the left side using perfect square trinomials and simplify the right side. ( y ) ( ) = Finally, divide through by the right hand side to make it. Reduce any fractions so the coefficients are completely in the denominat. ( y ) ( 1) = 1 6 5

17 Practice: Complete the square; find the center, change in and y, and codinates of the vertices and foci. 16 9y y 9 = 0 Asymptotes 95. The asymptotes of a hyperbola are a pair of intersecting. Δy ± Δ 96. The of the asymptotes will be. 97. The asymptotes will pass through the of the hyperbola. Δy Δ ( y k) =± ( h) 98. The of the asymptotes are. 99. The further the graph is from the, the closer it gets to the asymptotes The graph of the hyperbola will cross the asymptotes The equations of the asymptotes don't depend on which the hyperbola opens. Practice: Find the equations of the asymptotes of the hyperbola. y = ( y+ ) ( 3) = 1 4 5

18 Parabolas 10. Parabolas are easy to spot because both variables are present, but only one variable is. Standard Fm Vertical Ais of Symmetry = 4 py Hizontal Ais of Symmetry ( h) = 4 p( y k ) ( y k) = 4 p( h) y = 4 p ( ) = y ( y+ 1) = ( 3) 103. In the simplest fm of a parabola (without the h and k), the verte is at the The focus always lies the parabola is the distance from the verte to the focus and is called the focal length.

19 106. The same distance as the focus, but on the other side of the verte is a line called the The ais of symmetry passes through the and and is perpendicular to the The direction of the ais of symmetry is determined by which variable is squared The parabola will open in the positive direction (up right) if p is and in the negative direction (down left) if p is. Finding the Verte and Focal Length Consider the parabola + 8y= Get the term by itself. = 8y Fact the right hand side so the coefficient on the linear variable is. 1 = 8 y 11. The constant on the right hand side will be. So divide that by 4 to find p. 4p = 8 p = 113. This parabola has opens down because the is the linear variable and p is negative The verte is at (, ) and the focal length is Since the focus is units below the verte, the directri is units above the verte The equation of the directri is y =.

20 Completing the Square Consider the parabola y y 3 + 5= Determine which variable is. Move the constant and the linear variable to the other side. y y = If the coefficient on the squared variable isn't, then divide through by that value. In this case, we don't need to do that. Leave space f a third number on the left hand side. y y + = Take the linear coefficient and square it. Add that amount to both sides. y y + 1= the left side using perfect square trinomials and simplify the right side. Fact a constant out of the right side if the coefficient isn't one. ( y ) ( y ) 1 = = 3 3 Now it's in standard fm and 4p = 3, so 3 p = The parabola has a ais of symmetry because the is the linear variable. 1. The verte is at (, ). 13. The focus is 3/4 units to the of the verte. That puts it at 5, , The directri is 3/4 units to the of the verte. That puts it at = = 3 4 1

21 Practice: Find the verte, focus, and directri and give the direction the parabola opens. = y y y + 4 = 0 Definition of a parabola 15. A parabola is the set of all points in a plane from a fied point and a line. 16. The fied point is called the. 17. The line is called the. 18. Distances from a point to a line are always measured to the line. Finding the equation of the parabola from the graph 19. The first thing to do is determine the general fm of the parabola based on which direction it Since this graph opens to the, the general fm is. y = 4 p 131. Determine where the verte is and make substitutions into the equation if it's not at the igin. Here, the verte is at (, ), so replace with and y with. ( y ) = 4p( + ) + y 13. Find another on the parabola and use it f and y. Here the point (,0) is on the parabola. ( 0 ) = 4p( + )

22 133. Solve the equation f to find the focal length. ( ) 4= 4p 4 4= 16p 1 p = Double check to make sure the on p agrees with the direction the parabola opens. If the parabola opens down to the left, then p should be. If the parabola opens up to the right, then p should be Finally, substitute the value f p into the equation and. 1 4 = + ( y ) = 4 ( + ) ( y ) Practice: Find the equation of the following parabolas.

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