240 likes | 717 Views
The general equation for all conic sections is:. Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. where A, B, C, D, E and F represent constants.
E N D
The general equation for all conic sections is: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B, C, D, E and F represent constants.
The graph of a quadratic relation will be a parabola if the coefficientof the x2 term or y2term is zero (and the xy term is zero).
Parabola The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.
Parabola Focus Directrix The line is called the directrixand the point is called the focus.
Parabola Focus d1 Directrix d2 d1= d2 for any point (x, y) on the parabola
Parabola Axis of Symmetry Focus Vertex Directrix The line perpendicular to the directrix passing through the focus is the axis of symmetry. Thevertexis the point of intersection of the axis of symmetry with the parabola.
Parabola Focus p Directrix The distance from the focus to the vertex is called the focal length and is called p.
Parabola 4p Focus Directrix The chord through the focus that is parallel to the directrix is length |4p| and is called the focal width.
Standard Form for a Parabola Theequation x2 = 4pyis the standard form equation for a parabola with vertex at (0, 0) that opens upwards or downwards. If p > 0, the parabola opens up. If p < 0, the parabola opens down.
p > 0 p < 0 x2 = 4py x2 = 4py Directrix: y = -p Focus (0, p) Focus (0, p) Directrix: y = -p
Standard Form for a Parabola Theequation y2 = 4px is the standard form equation for a parabola with vertex at (0, 0) that opens right or left. If p > 0, the parabola opens right. If p < 0, the parabola opens left.
p > 0 p < 0 y2 = 4px y2 = 4px Directrix: x = -p Directrix: x = -p Focus (p, 0) Focus (p, 0)
Example 1: Find the focus, directrix and focal width of the parabola x2 = -2y Compare to x2 = 4py: x2 = -2y 4p = -2 p = - ½ Does it open up or down? Focus:(0, - ½) Directrix:y = - (- ½) = ½ Focal width: |4p| = 2
Example 2: Find an equation in standard form for the parabola whose directrix is the line x = 2 and whose focus is the point (-2, 0) • Create a quick sketch. We can tell it will open left. p = -2 y2 = 4px y2 = 4(-2)x y2 = -8x
Example 3: • Identify the focus, directrix and focal width of the parabola (x + 4)2 = -12(y + 1). The vertex is (-4, -1) and p = -3. Focus: (h, k + p) = (-4, -1 + -3) = (-4, -4) Directrix: y = k – p = -1 – (-3) y = 2 Focal width: |4p| = |4(-3)| = 12
Example 4: • Find the standard form of the equation of the parabola with vertex (3, 4) and focus (5, 4). Let’s make a quick sketch. Since it opens to the right, use the form (y – k)2 = 4p(x – h) The focal length is p = 2. (y – 4)2 = 8(x – 3)