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**The Beginning of Chapter 8:**

Conic Sections (8.1a) Parabolas!!!

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**V Imagine two non-perpendicular lines intersecting at point V.**

Rotating one of the lines (the generator) around the other (the axis) yields a pair of right circular cones… A conic section is formed by the intersection of a plane and these cones. V Basics: Parabola, Ellipse, Hyperbola Degenerates: Point, Line, Intersecting Lines Check out the diagram on p.631…

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**Conic Sections All conic sections can be defined algebraically as the**

graphs of second-degree (quadratic) equations in two variables…in the form: where A, B, and C are not all zero. Today, we focus on parabolas!!!

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Definition: Parabola A parabola is the set of all points in a plane equidistant from a particular line (the directrix) and a particular point (the focus) in the plane. Point on the parabola Axis Dist. to focus Dist. to directrix Focus Vertex Directrix

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**Let’s equate these two distances:**

Deriving the equation of a parabola Focus F(0, p) p P(x, y) p Let’s equate these two distances: Directrix: y = –p D(x, –p)

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**Deriving the equation of a parabola**

Standard form of the equation of an up- or down-opening parabola. If p > 0, it opens up, if p < 0, it opens down.

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**Deriving the equation of a parabola**

Focus F(0, p) p P(x, y) p Directrix: y = –p D(x, –p) The value p is the focal length of the parabola. A segment with endpoints on a parabola is a chord. The value |4p| is the focal width.

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**Deriving the equation of a parabola**

Focus F(0, p) p P(x, y) p Directrix: y = –p D(x, –p) Parabolas that open right or left are inverse relations of the upward or downward opening parabolas…standard form:

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**Parabolas with Vertex (0, 0)**

Standard Equation Opens Upward or downward To the right or to the left Focus Directrix Axis y-axis x-axis Focal Length Focal Width

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**Guided Practice Focus: (0, –1/2), Directrix: y = 1/2, Focal Width: 2**

Find the focus, the directrix, and the focal width of the given parabola. Then, graph the parabola by hand. Focus: (0, –1/2), Directrix: y = 1/2, Focal Width: 2

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Guided Practice Find an equation in standard form for the parabola whose directrix is the line x = 2 and whose focus is the point (–2, 0). Would a graph help??? y = –8x 2

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Guided Practice Find an equation in standard form for the parabola whose vertex is (0, 0), opens downward, and has a focal width of 4. Would a graph help??? 2 x = – 4y

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**Whiteboard Practice … Standard Form:**

Find an equation in standard form for the parabola with vertex (0, 0), opening upward, with focal width = 3. (since parabola opens upward) Standard Form:

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**Translations of Parabolas**

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**We have only considered parabolas with the vertex on the**

origin………………… what happens when it’s not??? V F (h, k + p) (h, k) F (h + p, k) V (h, k) Such translations do not change the focal length, the focal width, or the direction the parabola opens!!!

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**Parabolas with Vertex (h, k)**

Standard Equation Opens Upward or downward Focus Directrix Axis Focal Length Focal Width

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**Parabolas with Vertex (h, k)**

Standard Equation Opens To the right or to the left Focus Directrix Axis Focal Length Focal Width

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Practice Problems Find the standard form of the equation for the parabola with vertex (3, 4) and focus (5, 4). Which general equation do we use? What are the values of h and k? Now, how do we find p?

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**Practice Problems Use a function grapher to graph the given parabola.**

First, we must solve for y!!! Now, plug these two equations into your calculator!!!

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Practice Problems Prove that the graph of the given equation is a parabola, then find its vertex, focus, and directrix. We need to complete the square… The CTS step!!! We have h = 2, k = –1, and p = 6/4 = 1.5 Vertex: (2, –1), Focus: (3.5, –1), Directrix: x = 0.5

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**Practice Problems Standard Form:**

Find an equation in standard form for the parabola that satisfies the given conditions. Vertex (–3, 3), opens downward, focal width = 20 (since parabola opens downward) Standard Form:

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**Practice Problems Standard Form:**

Find an equation in standard form for the parabola that satisfies the given conditions. Vertex (2, 3), opens to the right, focal width = 5 (since parabola opens to the right) Standard Form:

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Find the distance between (-4, 2) and (6, -3). Find the midpoint of the segment connecting (3, -2) and (4, 5).

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