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Conic Sections: The parabola

Conic Sections: The parabola. Colleen Beaudoin For FCIT. Parabola. Review: The geometric definition relies on a cone and a plane intersecting it. Algebraic definition: All points that are equidistant from a given line (the directrix)and a fixed point not on the directrix (the focus).

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Conic Sections: The parabola

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  1. Conic Sections:The parabola Colleen Beaudoin For FCIT

  2. Parabola • Review: The geometric definition relies on a cone and a plane intersecting it. • Algebraic definition: All points that are equidistant from a given line (the directrix)and a fixed point not on the directrix (the focus)

  3. Parabola Any point on the parabola is equidistant to the focus and the directrix. Example: Point A: d1=d2 Point B: d1=d2 Focus Directrix A d2 B d1 d1 d2 x y

  4. Items referenced on the graph of a parabola: Vertex Focus Directrix Axis of Symmetry x y

  5. Facts: Parabola Equations • One variable is squared and one is not. (How does this differ from linear equations?) • There are many ways the equation of a parabola can be written. We will get the quadratic part (variable that is squared) on the left of the equal sign and the linear part (variable is to the first power) on the right of the equal sign. • Equation: (x - h)2 = c(y – k) OR (y - k)2 = c(x – h)

  6. (x - h)2 = c(y – k) OR (y - k)2 = c(x – h)where the vertex is at (h,k) and |c| is the width at the focus To graph: 1. Put in standard form (above) – squared term on left 2. Decide which way the parabola opens. Look at the right side. If y: + c → opens up If y: - c → opens down If x: + c → opens right If x: - c → opens left

  7. (x - h)2 = c(y – k) OR (y - k)2 = c(x – h)where the vertex is at (h,k) and |c| is the width at the focus To graph: 3. Plot the vertex (h,k) Note what happens to the signs. 4. Plot the focus: move │¼ c │ from the vertex in the direction that the parabola opens. Mark with an f. 5. Draw the directrix: │¼ c │ from the vertex in the opposite direction of the focus (Remember that the directrix is a line.)

  8. (x - h)2 = c(y – k) OR (y - k)2 = c(x – h)where the vertex is at (h,k) and |c| is the width at the focus To graph: 6. Plot the endpoints of the latus rectum/focal chord (width at the focus). The width is the │c│ at the focus. 7. Sketch the parabola by going through the vertex and the endpoints of the latus rectum. (Be sure to extend the curve and put arrows.) 8. Identify the axis of symmetry. (The line that goes through the vertex dividing the parabola in half.)

  9. Exp. 1: Graph (x - 5)2 = 12(y – 6) To graph: 1. Put in standard form– squared term on left Done 2. Decide which way the parabola opens. Look at the right side. If y: + c → opens up If y: - c → opens down If x: + c → opens right If x: - c → opens left Up because y is on the right and 12 is positive

  10. Exp. 1: Graph (x - 5)2 = 12(y – 6) To graph: 3. Plot the vertex (h,k) Note what happens to the signs. (5,6) 4. Plot the focus: move │¼ c │ from the vertex in the direction that the parabola opens. Mark with an f. (5,9): found by moving up 3 from the vertex 5. Draw the directrix: │¼ c │ from the vertex in the opposite direction of the focus (Remember that the directrix is a line.) y = 3: found by moving down 3 from the vertex

  11. Exp. 1: Graph (x - 5)2 = 12(y – 6) To graph: 6. Plot the endpoints of the latus rectum/focal chord (width at the focus). The width is the │c│ at the focus. L.R. = 12 with endpoints at (-1,9) & (11,9) 7. Sketch the parabola by going through the vertex and the endpoints of the latus rectum. (Be sure to extend the curve and put arrows.) 8. Identify the axis of symmetry. (The line that goes through the vertex dividing the parabola in half.) x = 5

  12. Exp. 1: Graph (x - 5)2 = 12(y – 6) Vertex: (5,6) Focus: (5,9) Directrix: y = 3 L.R.: 12 Axis: x = 5 f

  13. Exp. 2: Graph (y + 3)2 = -4(x – 2) To graph: 1. Put in standard form Done 2. Decide which way the parabola opens. Left because x is on the right and 4 is negative 3. Plot the vertex (h,k) (2,-3) 4. Plot the focus: (1,-3): found by moving left 1 from the vertex 5. Draw the directrix: x = 3: found by moving right 1 from the vertex

  14. Exp. 2: Graph (y + 3)2 = -4(x – 2) To graph: 6. Plot the endpoints of the latus rectum L.R. = 4 with endpoints at (1,-1) & (1,-5) 7. Sketch the parabola 8. Identify the axis of symmetry. y = -3

  15. Exp. 2: Graph (y + 3)2 = -4(x – 2) Vertex: (2,-3) Focus: (1,-3) Directrix: x = 3 L.R.: 4 Axis: y = -3 f

  16. Exp. 1: Graph x = y2 - 4y + 1 • What’s the first step? Put in standard form. y2 - 4y + 1 = x y2 - 4y + 4 = x - 1 + 4 Complete the square. (y – 2)2 = x + 3 (y – 2)2 = 1(x + 3) • Now you try graphing the parabola and labeling all the parts.

  17. Challenge Question Given the following information, write the equation of the parabola. Vertex is (0,0) and Focus is at (0,2)

  18. Review • How can you tell the graph of an equation will be a parabola? • What’s the standard form of a parabola? • What are the steps for graphing a parabola? • What are common errors people make when graphing parabolas?

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