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5.2 Parabolas 1 The following are several definitions necessary for the understanding of parabolas. 1.) Parabola - A parabola is the set of all points that are equidistant from a fixed line called the directrix and a fixed point called the focus (not on the line.) 2.) Axis of symmetry - A line passing through the focus and being perpendicular to the directrix. Focus ● P’P’ directrix Axis of symmetry ● ● V P ● 3.) Vertex - The point at which a parabola makes its sharpest turn. The vertex is halfway between the directrix and the focuspointparabola directrixfocus The distance from a point P to the focus equals the distance from the point P to the directrix (distance to P’). Next Slide

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5.2 Parabolas 2 F(0,p) ● y ● x P(x,y) y=-p P’(x,-p) ● Summary: The graph of x 2 = 4py has the following: Vertex: (origin) Focus: (0,p) Directrix: y=–p y-axis symmetry A line segment that contains the focus and whose endpoints are on the parabola is called a focal chord. The specific focal chord that is parallel to the directrix, we shall call the primary focal chord. The entire length of the primary focal chord is |4p| units. This will be useful when graphing parabolas. The line QP is the primary focal chord. y ● F(0,p) ● x P(x,p) Q P’(x,-p) ● ● y ● F(p,0) ● x P(p,y) Q(P,-y) P’(-p,y) ● ●

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5.2 Parabolas 3 Procedure: To find the focus and directrix of the parabola x 2 = 4py and sketch its graph. Step 1.Compare to the standard form x 2 = 4py to find p and the focus. If p>0, the parabola opens upward. If p<0, the parabola opens downward. The focus is at (0,p). Step 2.Find the directrix. The equation for the directrix is y =–p. Step 3.Find the endpoints of the primary focal chord. The length of the primary focal chord is |4p| units long. Therefore, the endpoints of the primary focal chord are (2p,p) and (-2p,p), (or |2p| units to the left and right of the focus). Example 1.Find the focus, directrix and focal chord endpoints of the parabola x 2 = 8y, and sketch its graph. Step 1.Compare x 2 = 8y to the standard form x 2 = 4py to find p and the focus. Step 3.Find the endpoints of the primary focal chord. The length of the primary focal chord is |4p| units long. Then |4(2)| = 8 units long. Therefore, the endpoints of the primary focal chord are (4,2) and (-4,2). F(0,2) ● y x Step 2.Find the directrix. The equation for the directrix is y =–p. The directrix for this parabola is y=–2. y=-2 Since p>0, the parabola opens upward and the focus is at (0,2). ● (4,2) ● (-4,2)

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5.2 Parabolas 4 x axis y axis focus: (0,3) directrix: y=-3 endpoints (6,3), (-6,3) Answer Your Turn Problem #1 Find the focus, directrix and focal chord endpoints of the parabola x 2 = 12y and sketch its graph.

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5.2 Parabolas 5 Parabolas with Vertices Not at the Origin Opening Up or Down x=h vertex (h,k) y=k p focus (h, k+p) This is similar to the equation of a circle not at the origin, (x h) 2 + (y k) 2 = r 2. The graph of the following equation is a parabola that has it vertex at (h,k) and has the indicated focus, directrix and symmetry. Focus (h, k+p), directrix y=k p, line of symmetry x = h. The length of the focal chord is |4p| units. Therefore the focal points are |2p| units to the left and right of the focus point. Next Slide

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5.2 Parabolas 6 Example 2. Find the vertex, focus, directrix, focal chord endpoints of the given parabola and sketch its graph. Vertex, (h,k) = (3,-1), and p=2. Solution: x axis y axis Focus (h, k+p), = (3, 1+2) = (3, 1) directrix y=k p, so y= 1 2 or y= 3. The last step before sketching the curve is to find the endpoints of the focal chord. The endpoints are |2p| or |2(2)|= 4 units to the left and right of the focus, which is (-1,1) and (7,1). Now we can sketch the curve. (3,1) (7,1)(-1,1) (3,-1) y=-3 Your Turn Problem #2 Find the vertex, focus, directrix, focal chord endpoints of the parabola and sketch its graph. vertex: (-2,1) focus: (-2,4) directrix: y=-2 endpoints (-8,4), (4,4) x axis y axis (4,4)(-8,4) (-2,4) (-2,1) y=-2 Answer:

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5.2 Parabolas 7 Vertex, (h,k) = (4,0), and p=-2. Example 3. Find the vertex, focus, directrix, focal chord endpoints of the given parabola and sketch its graph. Solution: x axis y axis Focus (h, k+p), = (4, 0 - 2) = (4, -2) directrix y=k p, so y=0 ( 2) or y=2. The last step before sketching the curve is to find the endpoints of the focal chord. The endpoints are |2p| or |2(-2)|= 4 units to the left and right of the focus, which is (0,-2) and (8,-2). Now we can sketch the curve. (4,-2) (4,0) (8,-2) (0,-2) y=2 Your Turn Problem #3 Find the vertex, focus, directrix, focal chord endpoints of the parabola and sketch its graph. (-10,1/2) vertex: (-5,3) focus: (-5,1/2) directrix: y=5 1/2 endpoints (-10, 1/2), (0, 1/2) Answer: x axis y axis (0,1/2) (-5,3) (-5,1/2) y=5 1/2

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5.2 Parabolas 8 Summary: The graph of y 2 = 4px has the following: Vertex: (origin) Focus: (p,0) Directrix: x=–p x-axis symmetry F(p,0) ● y ● x P(x,y) x=-p P’(-p,y) ● Step 1.Compare to the standard form y 2 = 4px to find p and the focus. If p>0, the parabola opens to the right. If p<0, the parabola opens to the left. The focus is at (p,0). Step 2.Find the directrix. The equation for the directrix is x =–p. Procedure:To find the focus and directrix of the parabola y 2 = 4px and sketch its graph. Step 3.Find the endpoints of the primary focal chord. The length of the primary focal chord is |4p| units long. Therefore, the endpoints of the primary focal chord are (p,2p) and (p,-2p), (or |2p| units to above and below the focus.

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5.2 Parabolas 9 Example 4.Find the focus, directrix and focal chord endpoints of the parabola y 2 = 12x, and sketch its graph. Step 1.Compare y 2 = 12x to the standard form y 2 = 4px to find p and the focus. Step 2.Find the directrix. The equation for the directrix is x =–p. The directrix for this parabola is x=–3. Since p>0, the parabola opens to the right and the focus is at (3,0). Step 3.Find the endpoints of the primary focal chord. The endpoints are |2p| or |2(3)|= 6 units above and below the focus, which is (3,-6) and (3,6). F(3,0) ● y x x=-3 y 2 = 12x ● (3,6) ● (3,-6) x axis y axis focus: (-1,0) directrix: x=1 endpoints (-1,2), (-1,-2) Answer: Your Turn Problem #4 Find the focus, directrix and focal chord endpoints of the parabola y 2 = -4x and sketch its graph.

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5.2 Parabolas 10 Parabolas with Vertices Not at the Origin Left or Right x=h p vertex (h,k) y=k The graph of the following equation is a parabola that has it vertex at (h,k) and has the indicated focus, directrix and symmetry. Focus (h+p, k), directrix x= h p, line of symmetry y = k. The length of the focal chord is still |4p| units. Therefore the focal points are |2p| units above and below focus point. focus (h+p, k) Next Slide

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5.2 Parabolas 11 Example 5. Find the vertex, focus, directrix, focal chord endpoints of the given parabola and sketch its graph. Vertex, (h,k) = (-1,3), and p=2. Solution: x axis y axis Focus (h+p, k), = (-1+2, 3) = (1, 3) directrix x=h p, so x= 1 2 or x= 3. The last step before sketching the curve is to find the endpoints of the focal chord. The endpoints are |2p| or |2(2)|= 4 units to the left and right of the focus, which is (1,-1) and (1,7). Now we can sketch the curve. (1,3) (-1,3) (1,-1) (1,7) x=-3 Next Slide

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5.2 Parabolas 12 Your Turn Problem #5 Find the vertex, focus, directrix, focal chord endpoints of the parabola and sketch its graph. x axis y axis vertex: (3,4) focus: (2 1/2, 4) directrix: x=3 1/2 endpoints (2 1/2,5), (2 1/2,3) Answer

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5.2 Parabolas 13 Writing a Parabola in Standard Form: or Observe the left hand side (LHS) of each standard form. Both are the factored form of a perfect square trinomial where the leading coefficient is a 1. Observe the right hand side (RHS) of each standard form. Both contain the variable which is not squared. The RHS is factored so that the leading coefficient is a 1. 1.Write the variable which is squared along with any terms with the same variable on the LHS. 2. Write the variable which is not squared along with the constant term on the RHS. Next Slide 3.Use completing the square on the LHS to create a binomial squared. It is not necessary in this example because there is no x term. Factor out the coefficient of the y term on the RHS.

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5.2 Parabolas 14 Write the variable which is squared along with any terms with the same variable on the LHS. Write the variable which is not squared along with the constant term on the RHS. On the LHS, use completing the square to create a binomial squared. It is not necessary in this example because there is no x term. Factor out the coefficient of the y term on the RHS. From here, we could find the vertex, focus, directrix and graph. Your Turn Problem #6

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5.2 Parabolas 15 Write the variable which is squared along with any terms with the same variable on the LHS. Write the variable which is not squared along with the constant term on the RHS. On the LHS, use completing the square to create a binomial squared. Multiply the middle term by ½, then square it. Add that number to both sides. Again, from here, we could find the vertex, focus, directrix and graph. Write the LHS as a binomial squared. Simplify the RHS and factor out the leading coefficient. Your Turn Problem #7

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5.2 Parabolas 16 Example 8. Find the vertex, focus, directrix, focal chord endpoints of the given parabola and sketch its graph. Vertex, (h,k) = (-3,2), and p=2. Solution: x axis y axis Focus (h+p, k), = (-3+2, 2) = (-1, 2) directrix x=h p, so x= 3 2 or x= 5. The last step before sketching the curve is to find the endpoints of the focal chord. The endpoints are |2p| or |2(2)|= 4 units to the left and right of the focus, which is (-1,-1) and (-1,7). Now we can sketch the curve. Next Slide (-1,2)(-3,2) (-1,-2) (-1,6) x=-5 Rewrite in standard form.

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5.2 Parabolas 17 Your Turn Problem #8 Find the vertex, focus, directrix, focal chord endpoints of the parabola and sketch its graph. vertex: (1,3) focus: (4, 3) directrix: x=-2 endpoints (4,9), (4,-3) Answer x axis y axis (4,3) (1,3) (1,-1) (1,9) x=-2

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5.2 Parabolas 18 For the remainder of this section, we will no longer concentrate on graphing the parabola. We will instead concentrate on finding the equation of the parabola that satisfies certain conditions. Keep in mind the vertex is the midpoint of the focus and the closest point on the directrix. Finding the Equation of a Parabola ● Same distance Next Slide Although, we are not required to graph the parabola, it is definitely a good idea to make a rough sketch using the information given and your knowledge of parabolas. ● ● focus directrix vertex

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5.2 Parabolas 19 Example 9. Find the equation of the parabola that satisfies the given conditions: Focus (0,2), directrix y=–2 Solution: Determine direction of parabola by sketching focus and directrix. The vertex is the midpoint of the focus and the closest point on the directrix: (0,0) The focus is contained within the parabola. The parabolas direction is upward. ● vertex ● We need to find p. Because the parabola opens upward, p>0. If the parabola were to open downward, p<0. p is the distance from the vertex to the focus, p=2. vertex: (0,0) p=2 Since the parabola opens upward, we use the following equation: Replace h, k and p with the values found, simplify and set equal to zero. Your Turn Problem #9 Find the equation of the parabola that satisfies the given conditions: Focus (0,–3), directrix y=3

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5.2 Parabolas 20 Determine direction of parabola by sketching focus and directrix. The vertex is the midpoint of the focus and the closest point on the directrix: (5,2). Take the total distance, 6–(–2) = 8 and divide by 2 which equals 4. 4 is the distance from the focus to the vertex or the distance from the vertex to the directrix. The focus is contained within the parabola. This parabola opens downward. We need to find p. Because the parabola opens downward, p<0. p=–4 Since the parabola opens downward, we use the following equation: Replace h, k and p with the values found, simplify and set equal to zero. Example 10. Find the equation of the parabola that satisfies the given conditions: Focus (5,–2), directrix y=6 Solution: x axis y axis ● focus vertex ● (5,2) vertex: (5,2) p=–4 Your Turn Problem #10 Find the equation of the parabola that satisfies the given conditions: Focus (–4,1), directrix y=–3

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5.2 Parabolas 21 Example 11. Find the equation of the parabola that satisfies the given conditions: Focus (8,0), directrix x=–2 Solution: Determine direction of parabola by sketching focus and directrix. The vertex is the midpoint of the focus and the closest point on the directrix: (3,0). Take the total distance, 8–(–2) = 10 and divide by 2 which equals 5. 5 is the distance from the focus to the vertex or the distance from the vertex to the directrix. The focus is contained within the parabola. This parabola opens to the right. We need to find p. Because the parabola opens to the right, p>0. p=5 Since the parabola opens to the right, we use the following equation: Replace h, k and p with the values found, simplify and set equal to zero. x axisy axis ● focus vertex: (3,0) p=5 vertex ● ( 3,0) Your Turn Problem #11 Find the equation of the parabola that satisfies the given conditions: Focus (2,0), directrix x=6

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5.2 Parabolas 22 Example 12. Find the equation of the parabola that satisfies the given conditions: Focus (–4,5), directrix x=2 Solution: Determine direction of parabola by sketching focus and directrix. The vertex is the midpoint of the focus and the closest point on the directrix: (–1,5). Take the total distance, 2–(–4) = 6 and divide by 2 which equals 3. 3 is the distance from the focus to the vertex or the distance from the vertex to the directrix. The focus is contained within the parabola. This parabola opens to the left. We need to find p. Because the parabola opens to the left, p<0. p=–3 Since the parabola opens to the right, we use the following equation: Replace h, k and p with the values found, simplify and set equal to zero. x axis y axis ● focus vertex: (–1,5) p=–3 vertex ● (–1,5) Your Turn Problem #12 Find the equation of the parabola that satisfies the given conditions: Focus (6,–6), directrix x=2 The End B.R

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