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8.2 Graph and Write Equations of Parabolas

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

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8.2 Graph and Write Equations of Parabolas

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  1. 8.2 Graph and Write Equations of Parabolas Where is the focus and directrix compared to the vertex? How do you know what direction a parabola opens? How do you write the equation of a parabola given the focus/directrix? What is the general equation for a parabola?

  2. and a fixed line, called the directrix. Parabolas A parabola is defined in terms of a fixed point, called the focus, focus 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. directrix axis of symmetry

  3. y P(x, y) F(0, p) x O y = –p D(x, –p) Horizontal Directrix Standard Equation of a parabola with its vertex at the origin is x2 = 4py p > 0: opens upward p < 0: opens downward focus: (0, p) directrix: y = –p axis of symmetry: y-axis

  4. y D(x, –p) P(x, y) x O F(p, 0) x = –p Vertical Directrix Standard Equation of a parabola with its vertex at the origin is y2= 4px p > 0: opens right p < 0: opens left focus: (p, 0) directrix: x = –p axis of symmetry: x-axis

  5. 4 2 -2 -4 2 4 -2 -4 Example 1 Graph . Label the vertex, focus, and directrix. y2 = 4px Identify p. y2 = 4(1)x So, p = 1 Since p > 0, the parabola opens to the right. Vertex: (0,0) Focus: (1,0) Directrix: x = -1

  6. 4 2 -2 -4 2 4 -2 -4 Example 1 Graph . Label the vertex, focus, and directrix. Y2 = 4x Use a table to sketch a graph 0 2 4 -2 -4 0 1 4 1 4

  7. STEP 2 Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4pxwhere p = – 2. The focus is (p, 0), or (– 2, 0). The directrix is x = – p, or x = 2. Because yis squared, the axis of symmetry is the x - axis. 18 STEP 1 Rewrite the equation in standard form. x = – Graphx= –⅛y 2. Identify the focus, directrix, and axis of symmetry. SOLUTION Write original equation. – 8x = y 2 Multiply each side by – 8.

  8. 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.

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

  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. 2.45 4.24 4.90 5.48 2.45

  11. STEP 2 equation 14 14 Rewrite the equation in standard form. STEP 1 y = – x 2 Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola. 3. y = – x 2 SOLUTION Write original equation. Multiply each side by – 4. – 4y = x 2 directrix axis of symmetry focus x 2 = – 4 0, –1 y = 1 Verticalx = 0

  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 y - values. y x 4 2 2.83 3.46 4.47

  13. Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola. STEP 2 equation 4. x = – y 2 0, x = – Y 2 = 4 x 13 Rewrite the equation in standard form. STEP 1 3 3 3 4 4 4 13 Y 2 x = SOLUTION Write original equation. Multiply each side by 3. 3x = y 2 directrix axis of symmetry focus Horizontaly = 0

  14. 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. y x 3.46 1.73 2.45 3 3.87

  15. 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 x2=4(6)y x2 = 24y

  16. Write an equation of the parabola shown. 3 2 32 The graph shows that the vertex is (0, 0) and the directrix is y = – p = for pin the standard form of the equation of a parabola. – ( )y 32 x2= 4• Substitute for p SOLUTION x2 = 4py Standard form, vertical axis of symmetry x 2 = 6y Simplify.

  17. Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus. 5. Directrix: y = 2 SOLUTION x 2 = 4py Standard form, vertical axis of symmetry x 2 =4 (–2)y Substitute –2 for p x 2 = – 8y Simplify.

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

  19. The EuroDish, developed to provide electricity in remote areas, uses a parabolic reflector to concentrate sunlight onto a high-efficiency engine located at the reflector’s focus. The sunlight heats helium to 650°C to power the engine. Solar Energy •Write an equation for the EuroDish’s cross section with its vertex at (0, 0). •How deep is the dish? Read more on page 498

  20. 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? x2, graph opens up or down, y2, 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? x2= 4py (opens up [p>0] or down [p<0]), y2 = 4px (opens right [p>0] or left [p<0])

  21. 8.2 Assignment p. 499, 3-17 odd, 27-33 odd, 39-45 odd

  22. 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?

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

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

  25. 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 (x − h)2 = 4p(y − k) vertex: (-3,3) h = -3, k = 3 p = -1 (x + 3)2 = 4(−1)(y − 3)

  26. Graph(x – 2)2 = 8 (y + 3). SOLUTION 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 STEP 1 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 STEP 2 STEP 3 Draw a curve through the points.

  27. 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)

  28. Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3). SOLUTION 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 1 Identify h and k. The vertex is at (– 2, 3), so h = –2 and k = 3. STEP 2 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).

  29. Write the standard form of a parabola with vertex at (3, – 1) and focus at (3, 2). 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 (x – h)2 = 4p(y– k) where p > 0. STEP 2 Identify hand 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 isp | = | – 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)

  30. 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)

  31. 8.2 Assignment day 2 p. 499, 26-32 even, 40-46 even p. 531, 3, 9, 15-16

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