Parabolas PowerPoint.pptx 9.2 - parabolas 1.ppt
- 1. Parabolas
- 2. Parabolas V Things you should already know about a parabola. Forms of equations y = a(x – h)2 + k opens up if a is positive opens down if a is negative vertex is (h, k) y = ax2 + bx + c opens up if a is positive opens down if a is negative vertex is Thus far in this course we have studied parabolas that are vertical - that is, they open up or down and the axis of symmetry is vertical f
- 3. A parabola is a set of points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus. For any point Q that is on the parabola, d2 = d1 Directrix Focus Q d1 d2
- 4. In this unit we will also study parabolas that are horizontal – that is, they open right or left and the axis of symmetry is horizontal In these equations it is the y-variable that is squared. V x = a(y – k)2 + h
- 5. Standard Form Equation of a Parabola Horizontal Parabola Vertical Parabola Vertex: (h, k) If 4p > 0, opens right If 4p < 0, opens left The directrix is vertical the vertex is midway between the focus and directrix (y – k)2 = 4p(x – h) (x – h)2 = 4p(y – k) Vertex: (h, k) If 4p > 0, opens up If 4p < 0, opens down The directrix is horizontal and the vertex is midway between the focus and directrix Remember: |p| is the distance from the vertex to the focus
- 6. Our parabola may have horizontal and/or vertical transformations. This would translate the vertex from the origin to some other place. The equations for these parabolas are the same but h is the horizontal shift and k the vertical shift: opens up The vertex will be at (h, k) opens down opens right opens left (𝒙−𝒉)𝟐 =𝟒𝒑(𝒚 −𝒌) (𝒙−𝒉)𝟐 =−𝟒 𝒑(𝒚 −𝒌) (𝒚 −𝒌)𝟐 =−𝟒 𝒑(𝒙−𝒉) (𝒚 −𝒌)𝟐 =𝟒𝒑(𝒙 −𝒉)
- 7. Graphing a Parabola 1. (y + 3)2 = 4(x + 1) Find the vertex, focus and directrix. Then sketch the graph of the parabola Vertex: __________ The parabola is ________________ and opens to the ________. 4p = 4 p = 1 F V Focus: ___________ Directrix:___________ (–1, –3) horizontal right Find p. (0, –3) x = –2 Why? The y-term is being squared. The 4 is positive. x y
- 8. |p| = 3 2. Find the standard form of the equation of the parabola given: the focus is (2, 4) and the directrix is x = – 4 Equation: (y – k)2 = 4p(x – h) x y V The ____________ is midway between the ________ and ___________, so the __________ is _________. vertex vertex focus directrix (–1, 4) The directrix is ______________ so the parabola must be ___________ and since the focus is always inside the parabola, it must open to the ________. vertical horizontal right 4(3) = 12 (y – __)2 = ___(x – ___) 4 12 1 Equation:
- 9. |p| = 2 3. Find the standard form of the equation of the parabola given: the vertex is (2, –3) and focus is (2, –5) F Equation: (x – h)2 = 4p(y – k) V The vertex is midway between a ___________ parabola that directrix for this parabola is _________. vertical down 4(2) = 8 Because of the location of the vertex and focus this must be the focus and directrix, so the opens _______. y = –1 (x – __)2 = ___(y – ___) 2 8 – 3 x y Equation:
- 10. 4. The equation we are given may not be in standard form and we'll have to do some algebraic manipulation to get it that way. (you did this with circles). Since y is squared, we'll complete the square on the y's and get the x term to other side. middle coefficient divided by 2 and squared 1 1 must add to this side too to keep equation = Now we have it in standard form we can find the vertex, focus, directrix and graph.
- 11. Converting an Equation Directrix: _______ y2 – 2y + 12x – 35 = 0 5. Convert the equation to standard form Find the vertex, focus, and directrix y2 – 2y + ___ = –12x + 35 + ___ 1 (y – 1)2 = –12(x – 3) The parabola is ___________ and opens _______. Vertex: ________ 4p = –12 p = –3 F V Focus: _________ (y – ___)2 = –12x + _____ 1 1 36 (3, 1) (0, 1) x = 6 horizontal left x y