Lesson26_students_

# Lesson26_students_ - y We need to solve for p which we know...

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MA 154 Lesson 26 Delworth Section 11.1 Parabolas 1 In this lesson, we will find equations of parabolas with some given conditions. Find an equation of the parabola that satisfies the given conditions. Hint: Sketches are always helpful. Ex 1) Vertex V(1, 2) Ex 2) Vertex V(3, 2) Opening Down Opening Left Passing through the point P(3, -1) Passing through the point (0, 5) Ex 3) Focus F(3, 0) Ex 4) Focus F(0, –6) Directrix x = –1 Directrix y = 2 Opening down means this is a vertical parabola (the x is the squared variable). We know values for h, k, x, and

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Unformatted text preview: y . We need to solve for p , which we know should be negative. Remember, the distance between the focus and the directrix is 2 p . MA 154 Lesson 26 Delworth Section 11.1 Parabolas 2 Ex 5) Vertex V(–3, 2) Ex 6) Vertex V(–1, –3) Directrix x = 3 Focus F(–1, 2) Ex 6) Vertex at the origin Ex 7) Vertex (3, –2) Symmetric to the y-axis Axis parallel to the x-axis Passing through the point P(6, 3) y-intercept 1 Distance between the vertex and directrix is p . A sketch will make clear the direction of opening. Basic form of equation is 2 4 x py =...
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## This note was uploaded on 03/30/2012 for the course MA 154 taught by Professor Delworth during the Spring '08 term at Purdue.

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Lesson26_students_ - y We need to solve for p which we know...

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