1 / 11

Sullivan Algebra and Trigonometry: Section 10.2 The Parabola

Sullivan Algebra and Trigonometry: Section 10.2 The Parabola. Objectives of this Section Find the Equation of a Parabola Graph Parabolas Discuss the Equation of a Parabola Work With Parabolas with Vertex at (h,k).

bernadette-freddy
Download Presentation

Sullivan Algebra and Trigonometry: Section 10.2 The Parabola

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Sullivan Algebra and Trigonometry: Section 10.2The Parabola • Objectives of this Section • Find the Equation of a Parabola • Graph Parabolas • Discuss the Equation of a Parabola • Work With Parabolas with Vertex at (h,k)

  2. A parabola is defined as the collection of all points P in the plane that are the same distance from a fixed point F as they are from a fixed line D. The point F is called the focus of the parabola, and the line D is its directrix. As a result, a parabola is the set of points P for which: d(F,P) = d(P,D) The equation of a parabola with vertex at (0,0), focus at (a,0), and directrix x = -a is y2 = 4ax

  3. d(P,D) P D: x = -a d(F,P) F = (a,0) d(F,P) = d(P,D)

  4. (1,4) x = -4 (4,0) Find the equation of a parabola with vertex at (0,0) and focus at (4,0). Graph the equation. The distance from the vertex to the focus is a = 4. So the equation of the parabola is y2 = 16x

  5. Discuss the equation: y2 = 10x The equation is of the form y2 = 4ax, where 4a= 10, so a = 5/2. So, the graph of the equation is a parabola with vertex (0,0), a focus at the point (5/2, 0) and directrixx = -5/2

  6. Equations of a Parabola: Vertex at (0,0); Focus on Axis Vertex Focus Directrix Equation Description (0,0) (a,0) x = -ay2 = 4ax Parabola, symmetric on x axis, opens right (0,0) (-a,0) x = ay2 = -4ax Parabola, symmetric on x axis, opens left (0,0) (0,a) y = -ax2 = 4ay Parabola, symmetric on y axis, opens up (0,0) (0,-a) y = ax2 = -4ay Parabola, symmetric on y axis, opens down

  7. Find the equation of a parabola with focus (0, -3) and directrix the line y = 3. This parabola will have a vertex at (0,0), since that point is the midpoint between the directrix and the focus. Since the focus is on the negative y axis with a = 3, the equation of the parabola is x2 = -4(3)y or x2 = -12y

  8. Parabolas With Vertex at (h,k); Axis of Symmetry Parallel to a Coordinate Axis Vertex Focus Directrix Equation (h,k) (h+a, k) x = h - a (y - k)2 = 4a(x - h) (h,k) (h-a, k) x = h + a (y - k)2 = -4a(x - h) (h,k) (h, k+a) x = k - a (x - h)2 = 4a(y- k) (h,k) (h, k - a) x = k + a (x - h)2 = -4a(y- k)

  9. Find the equation of a parabola with vertex at (-2, 3) and focus at (0, 3). Graph the equation. The vertex and focus both lie on the horizontal line y = 3 (the axis of symmetry). This distance from the vertex to the focus is a = 2. Since the focus lies to the right of the vertex, the parabola opens to the right. The equation has the form: (y - k)2 = 4a(x - h) where (h,k) = (-2, 3) and a = 2 (y - 3)2 = 4(2)(x - (-2)) (y - 3)2 = 8(x + 2)

  10. (0, 7) D: x = -4 F = (0,3) Axis of Symmetry: y = 3 (0, -1) (y - 3)2 = 8(x + 2)

  11. Find the vertex and focus of the following parabola: x2 + 8x = 4y - 8 First, use the method of completing the square involving the variable x to rewrite the parabola. x2 + 8x + ____ = 4y - 8 + ____ 16 16 (x + 4)(x + 4) = 4y + 8 (x + 4)2 = 4(y + 2) The equation is of the form (x - h)2 = 4a(y- k) Vertex: (h,k) = (-4, -2); Focus: (h, k + a) = (-4, -1)

More Related