Conic sections- Parabola STEM TEACH
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The document provides the steps to write the equation of a parabola in standard form given certain information about its properties. It then works through two example problems: 1) Given the equation of a parabola in general form, it rewrites the equation in standard form and determines the vertex, focus, directrix, and axis of symmetry. 2) Given the vertex and directrix of a parabola, it derives the standard form equation that satisfies those conditions. 3) Given the height and base width of an arch with a parabolic shape, it illustrates the situation, determines the necessary values, and writes the standard form equation that models the arch's shape.
Conic sections- Parabola STEM TEACH
- 2. PARABOLA
- 3. Given the equation of the parabola 𝑥2 + 6𝑥 + 12𝑦 + 15 = 0 write the equation in standard form and determine the following: a. Vertex b. Focus c. Equation of the directrix d. Axis of symmetry. e. Sketch the graph.
- 4. Solution: Since the second degree is raised in x, we expect that the standard form follows (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘).
- 5. 𝑥2 + 6𝑥 + 12𝑦 + 15 = 0 First step: Isolate 𝑥2 + 6𝑥 𝑥2 + 6𝑥 = −12𝑦 − 15
- 6. 𝑥2 + 6𝑥 = −12𝑦 − 15 Second step: Undergo completing the square process by adding a constant 9 on both side of the equation. 𝑥2 + 6𝑥 + (_) = −12𝑦 − 15 𝑥2 + 6𝑥 + 9 = −12𝑦 − 15 + 9
- 7. 𝑥2 + 6𝑥 + 9 = −12𝑦 − 15 + 9 Third step: Write the expressions in factored form in both side of the equation 𝑥 + 3 2 = −12𝑦 − 6
- 8. 𝑥 + 3 2 = −12𝑦 − 6 Fourth step: Express in the final answer in this form (𝑥 − ℎ)2= 4𝑝(𝑦 − 𝑘). 𝑥 + 3 2 = −12 𝑦 + 1 2
- 9. 𝑥 + 3 2 = −12 𝑦 + 1 2 𝑥 + 3 2 = −4 3 𝑦 + 1 2
- 10. Since the equation 𝑥 + 3 2 = −4 3 𝑦 + 1 2 is already written in standard form we may now determine the following: a. Vertex: V(h,k) = (-3, -1/2) b. Focus: F = k – p = -1/2 – 3 = -6 c. Equation of the directrix: y = k + p, y = 2.5 d. Axis of symmetry: x = h, x = -3 e. Sketch the graph.
- 11. Find an equation for the parabola that satisfies the given conditions whose vertex at V(9, 2) and directrix at x = 5.
- 12. Find an equation for the parabola that satisfies the given conditions whose vertex at V(9, 2) and directrix at x = 5. Solution 1: In order to find the equation we must plot the vertex and sketch the directrix in the plane to know where the parabola is facing.
- 13. Based from this illustration, the vertex is located at right of the directrix hence we can say that the opening of the parabola is at the left. Hence, we will use (𝑦 − 𝑘)2= −4𝑝 𝑥 − ℎ to determine the equation.
- 14. Solution 2: Since the vertex is 3 units to left of the directrix, p therefore is -3 and V(9, 2). Substitute these values in the equation (𝑦 − 𝑘)2= 4𝑝 𝑥 − ℎ First step: substitute these values in the standard form. (𝑦 − 2)2 = 4(3) 𝑥 − 9
- 15. (𝑦 − 2)2 = 4(3) 𝑥 − 9 Second step: Expand the expression 𝑦2 + 4𝑦 + 4 = 12 𝑥 − 9
- 16. 𝑦2 + 4𝑦 + 4 = 12 𝑥 − 9 Third step: Express the final answer in general equation form. 𝑦2 + 4𝑦 + 4 = 12𝑥 + 108
- 17. 𝑦2 + 4𝑦 + 4 = 12𝑥 + 108 𝑦2 + 4𝑦 − 12𝑥 − 104 = 0 Therefore, equation for the parabola that satisfies the given conditions whose vertex at V(9, 2) and directrix at x = 5 is 𝑦2 + 4𝑦 − 12𝑥 − 104 = 0
- 18. An arch in a memorial arch, having a parabolic shape, has a height of 50 feet and a base width of 80 feet. Find the equation of the parabola which models this shape, using x-axis to represent the ground.
- 19. Solution: To answer this problem, its best to make an illustration and identify the given values and what is asked.
- 20. To illustrate the situation, the coordinates must be plotted on the cartesian plane. The height 50 ft must be the basis of the vertex with x =0. The base 80ft is the distance away from the center that gives a coordinates of (40, 0) and (-40, 0). Since these are the values we have, we will substitute these in the standard form (𝑥 − ℎ)2 = 4𝑝 𝑦 − 𝑘 because the graph opens downward.
- 21. Solution: Use V(0,50) and a point (40,0) to determine the equation (𝑥 − ℎ)2= 4𝑝 𝑦 − 𝑘 (𝑥 − ℎ)2= 4𝑝 𝑦 − 𝑘 First step: substitute these values in the standard form. (40 − 0)2 = 4𝑝 0 − 50
- 22. (40 − 0)2 = 4𝑝 0 − 50 Second step: Simplify the equation and solve for p 1600 = −200𝑝 𝑝 = −8
- 23. 𝑝 = −8 Third step: Use the vertex and the new obtained value of p to substitute the equation. (𝑥 − ℎ)2 = 4𝑝 𝑦 − 𝑘 (𝑥 − 0)2 = 4(−8) 𝑦 − 50
- 24. (𝑥 − 0)2 = 4(−8) 𝑦 − 50 Fourth step: Simplify the result and write the final answer in general form. 𝑥2 = −32𝑦 + 1600 𝒙 𝟐 + 𝟑𝟐𝒚 − 𝟏𝟔𝟎𝟎 = 𝟎 The equation of the parabola which models this shape is 𝒙 𝟐 + 𝟑𝟐𝒚 − 𝟏𝟔𝟎𝟎 = 𝟎