1. Objectives: You will be able to define parametric equations, graph curves parametrically, and solve application problems using parametric equations. Agenda: 1.Do Now 2. Graphing parabolas with a directrix and focus. Precalculus April, 8, 2015

2. 8.1 Graph and Write Equations of Parabolas 8.1 Graph and Write Equations of Parabolas Where is the focus and directrix compared to the vertex?Where is the focus and directrix compared to the vertex? How do you know what direction a parabola opens?How do you know what direction a parabola opens? How do you write the equation of a parabola given the focus/directrix?How do you write the equation of a parabola given the focus/directrix? What is the general equation for a parabola?What is the general equation for a parabola?

3. Introduction to Conic Sections Conic Section-intersection of a plane and a cone. Circle, Ellipse, Hyperbola, Parabola http://www.math.odu.edu/cbii/calcani m/consec.avihttp://www.math.odu.edu/cbii/calcani m/consec.avi

4. Parabolas focus axis of symmetry directrix A parabola is defined in terms of a fixed point, called the focus, and a fixed line, called the directrix. A parabola is the set of all points P(x,y) in the plane whose distance to the focus equals its distance to the directrix.

5. Horizontal Directrix p > 0: opens upward focus: (0, p) directrix: y = –p axis of symmetry: y-axis x y D(x, –p) P(x, y) F(0, p) y = –p O p < 0: opens downward Standard Equation of a parabola with its vertex at the origin is x 2 = 4py Focal length(distance from vertex to the focus): p Vertex is halfway between directrix and focus

6. Vertical Directrix p > 0: opens right focus: (p, 0) directrix: x = –p axis of symmetry: x-axis p < 0: opens left Standard Equation of a parabola with its vertex at the origin is x y D(x, –p) P(x, y) F(p, 0) x = –p O y 2 = 4px Focal length: p

7. Example 1 Graph. Label the vertex, focus, and directrix. y 2 = 4px -4 -2 2 42 4 -4 -2 Identify p. So, p = 1 Since p > 0, the parabola opens to the right. Vertex: (0,0) Focus: (1,0) Directrix: x = -1 y 2 = 4(1)x

8. Example 1 Graph. Label the vertex, focus, and directrix. Y 2 = 4x -4 -2 2 42 4 -4 -2 Use a table to sketch a graph y x 0 0 2 1 4 4 -2 1 -4 4

9. SOLUTION STEP 1 Rewrite the equation in standard form. 1818 x = – Write original equation. Graph x = – ⅛ y 2. Identify the focus, directrix, and axis of symmetry. – 8x = y 2 Multiply each side by – 8. STEP 2 Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = – 2. The focus is (p, 0), or (– 2, 0). The directrix is x = – p, or x = 2. Because y is squared, the axis of symmetry is the x - axis.

10. STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values.

11. Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola. 1. Y 2 = –6x SOLUTION STEP 1 Rewrite the equation in standard form. Y 2 = 4 (– )x 3 2 STEP 2 Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = –. The focus is (p, 0), or (–, 0). The directrix is x = – p, or x =. Because y is squared, the axis of symmetry is the x - axis. 3 2 3 2 3 2

12. STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values. 2.45 4.244.905.48

13. Example 2 Write the standard equation of the parabola with its vertex at the origin and the directrix y = -6. Since the directrix is below the vertex, the parabola opens up Since y = -p and y = -6, p = 6 x 2 =4(6)y x 2 = 24y

14. SOLUTION The graph shows that the vertex is (0, 0) and the directrix is y = – p = for p in the standard form of the equation of a parabola. 3232 – x 2 = 4py Standard form, vertical axis of symmetry x 2 = 4 ( ) y3232 Substitute for p 3232 x 2 = 6y Simplify. Write an equation of the parabola shown.

15. Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus. x 2 = 4py Standard form, vertical axis of symmetry x 2 = 4 (3)y Substitute 3 for p x 2 = 12y Simplify. 8. Focus: (0, 3) SOLUTION

16. Where is the focus and directrix compared to vertex? The focus is a point on the line of symmetry and the directrix is a line below the vertex. The focus and directrix are equidistance from the vertex. How do you know what direction a parabola opens? x 2, graph opens up or down, y 2, graph opens right or left How do you write the equation of a parabola given the focus/directrix? Find the distance from the focus/directrix to the vertex (p value) and substitute into the equation. What is the general equation for a parabola? x 2 = 4py (opens up [p>0] or down [p 0] or left [p<0])

17. 8.2 Graph and Write Equations of Parabolas day 2 What does it mean if a parabola has a translated vertex? What general equations can you use for a parabola when the vertex has been translated?

18. Standard Equation of a Translated Parabola Vertical axis: vertex: (h, k) focus: (h, k + p) directrix: y = k – p axis of symmetry: x = h ( x − h) 2 = 4p(y − k)

19. Standard Equation of a Translated Parabola Horizontal axis: vertex: (h, k) focus: (h + p, k) directrix: x = h - p axis of symmetry: y = k (y − k) 2 = 4p(x − h)

20. Example 3 Write the standard equation of the parabola with a focus at F(-3,2) and directrix y = 4. Sketch the info. The parabola opens downward, so the equation is of the form vertex: (-3,3) h = -3, k = 3 p = -1 (x − h) 2 = 4p(y − k) (x + 3) 2 = 4(−1)(y − 3)

21. Graph (x – 2) 2 = 8 (y + 3). SOLUTION STEP 1 Compare the given equation to the standard form of an equation of a parabola. You can see that the graph is a parabola with vertex at (2, – 3), focus (2, – 1) and directrix y = – 5 Draw the parabola by making a table of value and plot y point. Because p > 0, he parabola open to the right. So use only points x- value x12345 y–2.875–3– 2.875–2.5– 1.875 STEP 3 Draw a curve through the points. STEP 2

22. Example 4 Write an equation of a parabola whose vertex is at ( −2,1) and whose focus is at (−3, 1). Begin by sketching the parabola. Because the parabola opens to the left, it has the form (y −k)2 = 4p(x − h) Find h and k: The vertex is at ( − 2,1) so h = − 2 and k = 1 Find p: The distance between the vertex (−2,1) and the focus (−3,1) by using the distance formula. p = −1 (y − 1) 2 = −4(x + 2)

23. Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3). SOLUTION STEP 1 Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (y – k) 2 = 4p(x – h) where p < 0. STEP 2 Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3. STEP 3 Find p. The vertex (– 2, 3) and focus (  4, 3) both lie on the line y = 3, so the distance between them is p | = | – 4 – (– 2) | = 2, and thus p = +2. Because p < 0, it follows that p = – 2, so 4p = – 8. The standard form of the equation is (y – 3) 2 = – 8(x + 2).

24. Write the standard form of a parabola with vertex at (3, – 1) and focus at (3, 2). SOLUTION STEP 1Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (x – h) 2 = 4p(y – k) where p > 0. STEP 2 Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1. STEP 3 Find p. The vertex (3, – 1) and focus (3, 2) both lie on the line x = 3, so the distance between them is p | = | – 2 – (– 1) | = 3, and thus p = + 3. Because p > 0, it follows that p = 3, so 4p = 12. The standard form of the equation is ( x – 3) 2 = 12(y + 1)

25. What does it mean if a parabola has a translated vertex? It means that the vertex of the parabola has been moved from (0,0) to (h,k). What general equations can you use for a parabola when the vertex has been translated? (y-k) 2 =4p(x-h) (x-h) 2 =4p(y-k)

26. 8.2 Assignment day 2 p. 499, 26-32 even, 40-46 even p. 531, 3, 9, 15-16