Sullivan Algebra and Trigonometry: Section 10.2 The Parabola

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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)

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

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

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 x = -4 (4,0) (1,4)

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 directrix x = -5/2

Equations of a Parabola: Vertex at (0,0); Focus on Axis Vertex Focus Directrix Equation Description (0,0) (a,0) x = -a y2 = 4ax Parabola, symmetric on x axis, opens right (0,0) (-a,0) x = a y2 = -4ax Parabola, symmetric on x axis, opens left (0,0) (0,a) y = -a x2 = 4ay Parabola, symmetric on y axis, opens up (0,0) (0,-a) y = a x2 = -4ay Parabola, symmetric on y axis, opens down

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

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)

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)

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

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)